P1-Allens Made Physics Theory {PART-2}

JEE-Physics GOLDEN  KEY  POINT • If  medium  (air)  is  also  moving  with  v   velocity  in  direction  of  source  to  observer.  Then  velocity  of  sound m relative  to  observer  will  be  v  ±  v  (–ve  sign,  if  v  is  opposite  to  sound  velocity).  So,    n'  =  n  v  vm  vo  m m  v  vm  vs  • If  medium  moves  in  a  direction  opposite  to  the  direction  of  propagation  of  sound,  then  n '   v  vm  vo  n  v  vm  vS   v • Source  in  motion  towards  the  observer.  Both  medium  and  observer  are  at  rest.  n '   v  vS  n So,  when  a  source  of  sound  approaches  a  stationary  observer,  the  apparent  frequency  is  more  than  the actual  frequency.  v • Source  in  motion  away  from  the  observer.  Both  medium  and  observer  are  at  rest.  n '   v  vS  n .  So,  when a  source  of  sound  moves  away  from  a  stationary  observer,  the  apparent  frequency  is  less  than  actual frequency. • Observer  in  motion  towards  the  source.  Both  medium  and  source  are  at  rest. n '   v  vo  n .  So,  when  v  observer  is  in  motion  towards  the  source,  the  apparent  frequency  is  more  than  the  actual  frequency. n '   v  vo  n  v  • Observer  in  motion  away  from  the  source.  Both  medium  and  source  are  at  rest.  .  So,  when observer  is  in  motion  away  from  the  source,  the  apparent  frequency  is  less  than  the  actual  frequency. • Both  source  and  observer  are  moving  away  from  each  other.  Medium  at  rest.  n '   v  vo  n  v  vS  DOPPLER'S  EFFECT  IN  REFLECTION  OF  SOUND  (ECHO) \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 When  the  sound  is  reflected  from  the  reflector  the  observer  receives  two  notes  one  directly  from  the  source  and other  from  the  reflector.  If  the  two  frequencies  are  different  then  superposition  of  these  waves  result  in  beats and  by  the  beat  frequency  we  can  calculate  speed  of  the  source. If  the  source  is  at  rest  and  reflector  is  moving  towards  the  source  with  speed  u, then  apparent  frequency  heard  by  reflector  n1   v  u n  v  Now  this  frequency  n   acts  as  a  source  so  that  apparent  frequency  received  by  observer  is 1 n2   v  n   v   v  u  n  v  u  n     v  u   v   v  u  v u 1 If  u  <<  v then n2  n 1  u 1  u  1  n 1  u2  n 1  2u v  v  v  v  Beat  frequency  n  =  n  –  n    2u n So  speed  of  the  source  u  v  n  2  v  2  n  26 E

JEE-Physics CONDITIONS  WHEN  DOPPLER'S  EFFECT  IS  NOT  OBSERVED  FOR  SOUND  WAVES • When  the  source  of  sound  and  observer  both  are  at  rest  then  Doppler  effect  is  not  observed. • When  the  source  and  observer  both  are  moving  with  same  velocity  in  same  direction. • When  the  source  and  observer  are  moving  mutually  in  perpendicular  directions. • When  the  medium  only  is  moving. • When  the  distance  between  the  source  and  observer  is  constant. Example When  both  source  and  observer  approach  each  other  with  a  speed  equal  to  the  half  the  speed  of  sound, then  determine  the  percentage  change  in  frequency  of  sound  as  detected  by  the  listener. Solution vv n' GFGGH v  v JKJIJ n  GFHGG  v JJJIK n  3n v  v Source 2 2   Observer 2 2 v 1 2 2 %  change  = n ' n  100  3n  n  100   =  2n  100   =  200  % nn n Example Two  trains  travelling  in  opposite  directions  at  126  km/hr  each,  cross  each  other  while  one  of  them  is  whistling. If  the  frequency  of  the  node  is  2.22  kHz  find  the  apparent  frequency  as  heard  by  an  observer  in  the  other train  : (a) Before  the  trains  cross  each  other (b) After  the  trains  have  crossed  each  other.  (v  sound =  335 m/sec) Solution 5 Here   v   =  126  ×    =  35  m/s 1 18 (i) In  this  situation    v       v 11 GFH KIJ FHG IKJn'  = v  v1 Observed  frequency v  v1  n   =  335  35  2220   =  2738  Hz 335  35 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 (ii) In  this  situation      v         v 1 1 Observed  frequency GHF IJK GFH KJIn'  = v  v1 v  v1  n   =  335  35  2220   =  1800  Hz 335  35 Example A  stationary  source  emits  sound  of  frequency  1200  Hz.  If  wind  blows  at  the  speed  of  0.1v,  deduce (a) The  change  in  the  frequency  for  a  stationary  observer  on  the  wind  side  of  the  source. (b) Report  the  calculations  for  the  case  when  there  is  no  wind  but  the  observer  moves  at  0.1v  speed  towards the  source.  (Given  :  velocity  of  sound  =  v) E 27

JEE-Physics Solution Medium  moves  in  the  direction  of  sound  propagation  i.e.  from  source  to  observer ( a) so  effective  velocity  of  sound veff  =  v  +  vm F I F Iv  vm  0 HG KJ GH KJsince  both  source  and  observer  are  at  rest   n' =  v  vm  0 n  =  v  0.1v n  =  n v  0.1v so  there  is  no  change  in  frequency FHG KJI(b) v  v0  v  0.1 v  When  observer  move  towards  source    n'  =  v n  =   v  n           =  1.1  n  =  1.1  ×  1200  Hz  =  1320  Hz Example A  bat  is  flitting  about  in  a  cave,  navigating  via  ultrasonic  beeps.  Assume  that  the  sound  emission  frequency  of  the bat  is  40  kHz.  During  one  fast  swoop  directly  toward  a  flat  wall  surface,  the  bat  is  moving  at  0.03  times  the speed  of  sound  in  air.  What  frequency  does  the  bat  hear  reflected  off  the  wall  ? Solution The  apparent  frequency  heard  by  the  bat  of  reflected  sound n '   v  v0    =   v  0.03v   40   =  1.03v  40 =  42.47  kHz  v  vs n  v  0.03v  0.97v   \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 28 E

JEE-Physics SOME WORKED OUT EXAMPLES Example#1 A particle of mass 50 g participates in two simple harmonic oscillations, simultaneously as given by x  = 1 10(cm) cos[80(s–1) t] and  x  = 5(cm) sin[(80(s–1) t + /6]. The amplitude of particle's oscillations  is given 2 by ‘A’. Find the value of A2 (in cm2). (A) 175 (B) 165 (C) 275 (D) 375 Solution Ans. (A) A A 2  A 2  2A1A2 cos   102  52  2  5 10  1  175   A2 = 175 1 2 2 Example#2 A sonometer wire resonates with a given tuning fork forming a standing wave with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire . When this mass is replaced by a mass ‘M’ kg, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. Find the value of M. (A) 25 (B) 20 (C) 15 (D) 10 Solution Ans. (A) 9g     f  2 /2 5/2= 5 Mg 9g Mg      f  2  f  2  2  M  5 3 /2= /2 3 5 3 Example#3 A steel wire of length 1 m and mass 0.1 kg and having a uniform cross-sectional area of 10–6 m2 is rigidly fixed at both ends. The temperature of the wire is lowered by 20°C. If the wire is vibrating in fundamental m o d e ,   f i n d   t h e   f r e q u e n c y   ( i n   H z ) . ( Y   =   2   ×   1 0 11   N / m 2,   ste   =   1 . 2 1   ×   1 0 –5 / ° C ) st ee l e l (A) 11 (B) 20 (C) 15 (D) 10 Solution Ans. (A) \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65     Y  T  T  YA   T  YA  48.4N ;  v  T 48.4 A    22m / s     0.1  1   for fundamental note        2m  f  v  22  11Hz 2 2 Example#4 A progressive wave on a string having linear mass density  is represented by  y  A sin  2 x  t  where y is    in mm. Find the total energy (in J) passing through origin from t = 0 to t = . 2 [Take :  = 3 × 10–2 kg/m; A = 1mm;  = 100 rad/sec;  = 16 cm] (A) 6 (B) 7 (C) 8 (D) 9 E 29

JEE-Physics Ans. (A) Solution Total energy  1 A 22   24 Example#5 Two tuning forks A and B lying on opposite sides of observer ‘O’ and of natural frequency 85 Hz move with velocity 10 m/s relative to stationary observer O. Fork A moves away from the observer while the fork B moves towards him. A wind with a speed 10 m/s is blowing in the direction of motion of fork A. Find the beat frequency measured b y   t he  ob se r ver  i n  Hz .  [ Ta ke  s p e ed  o f  so u nd  i n  ai r   as  3 40  m / s ] (A) 5 (B) 6 (C) 7 (D) 8 Solution Ans.  (A) fobserver for source 'A'  f0  v sound  v medium   33 ;  fobserver for source 'B'  f0  v sound  v medium   35 34 f0 34 f0  v sound  v medium  v source   v sound  v medium  v source       Beat frequency = f1  f2   3 5 3 3  f0  5  34  Example#6 If y  = 5 (mm) sint is equation of oscillation of source S  and y  = 5 (mm) sin(t + /6) be that of S 2 1 12 and it takes 1 sec and ½  sec for the transverse waves to reach point A from sources S  and S   respectively then 12 the resulting amplitude at point A, is A S1 S2 (A)  5 2  3 mm (B)  5 3  mm (C) 5 mm (D)  5 2 mm Ans. (C) Solution Wave originating at t =0 from S  reaches point A at t = 1. 1 1 Wave originating at t =  from S  reaches point A at t = 1. 22  2 2 So phase difference in these waves =   ; A =  A 1  A 2  2A1A2 cos   5 26 Example#7 A transverse wave, travelling along the positive x-axis, given by y=Asin(kx–t) is superposed with another wave travelling along the negative x-axis given by y=–Asin(kx+t). The point x=0 is (A) a node (B) an anitnode (C) neither a node nor an antinode (D) a node or antinode depending on t. Solution Ans.  (B) A t   x   = 0 ,   y   =  A si n   (–t)    and  y  =  – A si\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\nt; y1 y22A sin t (antinode) 2 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 1 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 Example#8 Figure shows a streched string of length L and pipes of length L, 2L, L/2 and L/2 in options (A), (B), (C) and (D) respectively. The string's tension is adjusted until the speed of waves on the string equals the speed of sound waves in air. The fundamental mode of oscillation is then set up on the string. In which pipe will the sound produced by the string cause resonance? L 30 E

JEE-Physics (A)  (B)   (C)  (D)  L 2L L/2 L/2 Solution Ans.  (B) Example#9 String  I  and  II  have  identical  lengths  and  linear  mass  densities,  but  string  I  is  under  greater  tension  than  string  II. The  accompanying  figure  shows  four  different  situations,  A  to  D,  in  which  standing  wave  patterns  exist  on  the two  strings.  In  which  situation  it  is  possible  that  strings  I  and  II  are  oscillating  at  the  same  resonant  frequency?   String  I String  II (A)  (B)  (C)  (D)  Solution Ans.  (C) Since  tension  in  I  >  tension  in  II    VI  >  V      Thus,  for  same  frequency,  I  >  II II Example#10 A  standing  wave  is  created  on  a  string  of  length  120  m  and  it  is  vibrating  in  6th  harmonic.  Maximum  possible amplititude  of  any  particle  is  10  cm  and  maximum  possible  velocity  will  be  10  cm/s.  Choose  the  correct statement.  Ans.  (A) (A)  Angular  wave  number  of  two  waves  will  be  . 20 (B)  Time  period  of  any  particle's  SHM  will  be  4  sec. (C)  Any  particle  will  have  same  kinetic  energy  as  potential  energy. (D)  Amplitude  of  interfering  waves  are  10  cm  each. Solution \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 6     120    40  k    A  v max   1  T  2  2  20 Example#11 Two  strings,  A  and  B,  of  lengths  4L  and  L  respectively  and  same  mass  M  each,  are  tied  together  to  form  a  knot 'O'  and  stretched  under  the  same  tension.  A  transverse  wave  pulse  is  sent  along  the  composite  string  from  the side  A,  as  shown  to  the  right.  Which  of  the  following  diagrams  correctly  shows  the  reflected  and  transmitted wave  pulses  near  the  knot  'O'? A OB E 31

JEE-Physics AO A O B (A)  B (B)  AO (D)  O AB (C)  B Solution Ans.  (A) The  wave  suffers  a  phase  difference  of    when  reflected  by  denser  medium. Example#12 Which  of  the  figures,  shows  the  pressure  difference  from  regular  atmospheric  pressure  for  an  organ  pipe  of length  L  closed  at  one  end,  corresponds  to  the  1st  overtone  for  the  pipe? (A)  (B)  (C)   (D)  Solution Ans.  (A) N A N For  pressure  standing  wave  antinode Note      A fundamental frequency first overtone Example#13 A man generates a symmetrical pulse in a string by moving his hand up and down. At  t = 0 the point in his hand moves downward. The pulse travels with speed of 3 m/s on the string & his hands passes 6 times in each second from the mean position. Then the point on the string at a distance 3m will reach its upper extreme first time at time t = (A) 1.25 sec. (B) 1 sec 13 (D) none Solution (C)  sec 12 Ans. (A) Frequency of wave   6  3  T  1 s ;    vT  3   1   1m \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 2 3  3  3 3T Total time taken =    1.25  sec 34 Example#14 Three progressive waves A, B and C are shown in figure. BA C With respect to wave A (A)    The  wave  C  lags  behind  in  phase  by  /2  and  B  leads  by  /2. (B)  The wave C leads in phase by  and B lags behind by  (C)    The  wave  C  leads  in  phase  by  /2  and  B  lags  behind  by  /2. (D)    The  wave  C  lags  behind  in  phase  by    and  B  leads  by  . Ans.  (A) 32 E

JEE-Physics Example#15 Following are equations of four waves : ( i )   y   =   a   s i n    t  x ( i i )   y   =   a   c o s    t  x     1 2 ( i i i )  z  =   a  s i n    t  x ( i v )   z   =   a   c o s    t  x 1     2 Which of the following statements is/are CORRECT? (A) On superposition of waves (i) and (iii), a travelling wave having amplitude a2 will be formed. (B) Superposition of waves (ii) and (iii) is not possible. (C) On superposition of waves (i) and (ii), a transverse stationary wave having maximum amplitude                a2 will be formed. (D) On superposition of waves (iii) and (iv), a transverse stationary wave will be formed. Solution Ans.  (AD) Superposition of waves (i) & (iii) will give travelling wave having amplitude of  a 2 {waves are along x-axis but particle displacements are along y & z-axis respectively} z1  z2  a    t  x  sin   t  x    sin  v    v     2  Example#16 T w o   me c h a n i c a l   w a v e s ,   y 1   =   2   s i n   2   ( 5 0   t     2x )   &   y =   4   s i n   2   ( a x   +   1 0 0  t )   p r o p a g a t e   i n   a   m e d i u m   w i t h 2 same speed. (A) The ratio of their intensities is 1: 16 (B) The ratio of their intensities is 1: 4 (C) The value of 'a' is 4 units (D) The value of 'a' is 2 units Solution Ans.  (AC) I  1 v2 A 2  and velocity =  2k \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 Example#17 Three simple harmonic waves, identical in frequency n and amplitude A moving in the same direction are superimposed in air in such a way, that the first, second and the third wave have the phase angles   ,    and 2 () respectively at a given point P in the superpositon. Then as the waves progress, the superposition will result in (A) a periodic, non-simple harmonic wave of amplitude 3A (B) a stationary simple harmonic wave of amplitude 3A (C) a simple harmonic progressive wave of amplitude A (D) the velocity of the superposed resultant wave will be the same as the velocity of each wave Solution Ans.  (CD) Since the first wave and the third wave moving in the same direction have the phase angles  and (+), they superpose with opposite phase at ever y point of the vibrating medium and thus cancel out each other, in displacement, velocity and acceleration. They, in effect, destroy each other out. Hence we are left with only the second wave which progresses as a simple harmonic wave of amplitude A. The velocity of this wave is the same as if it were moving alone. E 33

JEE-Physics Example#18 Two idetncial waves A and B are produced from the origin at different instants t  and t  along the positive x-axis, AB as shown in the figure. If the speed of wave is 5m/s then y(mm) AB 10 O 0.5 1 x(m) -5 -10 (A) the wavelength of the waves is 1m (B) the amplitude of the waves is 10 mm (C) the wave A leads B by 0.0167 s (D) the wave B leads A by 1.67 s Solution Ans.  (AB) Wavelength of the waves = 1m; Amplitude of the waves = 10 mm Example#19 A  progressive  wave  having  amplitude  5  m  and  wavelength  3  m.  If  the  maximum  average  velocity  of  particle  in half  time  period  is  5  m/s  and  wave  is  moving  in  the  positive  x-direction  then  find  which  may  be  the  correct equation(s)  of  the  wave?  [where  x  in  meter] (A)  5 sin  2 t  2 x  (B)  4 sin  t  2 x  3 cos  t  2 x  5 3   2 3  2 3 (C)  5 sin  t  2 x (D)  3 cos  2 t  2 x   4 sin  2 t  2 x  2 3  5 3   5 3 Solution Ans.  (BC)   3m   k  2  2  3 Maximum  displacement  in  half  time  period  =  2a  =  10  m So  maximum  average  velocity  =  10 5T  4 s    2  2   T T 4 2 2 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 Example#20 You  are  given  four  tuning  forks,  the  lowest  frequency  of  the  forks  is  300  Hz.  By  striking  two  tuning  forks  at  a time  any  of  1,  2,  3,  5,  7  &    8  Hz  beat  frequencies  are  heard.  The  possible  frequencies  of  the  other  three  forks are- (A)  301,302  &  307 (B)  300,304  &  307 (C)  301,  303  &  308 (D)  305,  307  &  308 Solution Ans.  (CD) 3 7 7 3 1 1 25 5 2 308 303 308 300 301 300 305 307 88 34 E

JEE-Physics Example#21 A  standing  wave  of  time  period  T  is  set  up  in  a  string  clamped  between  two  rigid  supports.  At t  =  0  antinode  is  at  its  maximum  displacement  2A. (A)  The  energy  density  of  a  node  is  equal  to  energy  density  of  an  antinode  for  the  first  time  at  t  =  T/4. (B)  The  energy  density  of  node  and  antinode  becomes  equal  after  T/2  second. T Ans.  (CD) (C)  The  displacement  of  the  particle  at  antinode  at  t  8   is  2A (D)  The  displacement  of  the  particle  at  node  is  zero Solution Equation  of  SHM  of  particle  who  is  at  antinode  is  y=2Asin  2  t   at  time  t  = T  T  8  y=  2Asin 4 =  2A;  Displacement  of  particle  at  note  is  always  zero. Example#22 Two  notes  A  and  B,  sounded  together,  produce  2  beats  per  sec.  Notes  B  and  C  sounded  together  produce 3  beats  per  sec.  The  notes  A  and  C  separately  produce  the  same  number  of  beats  with  a  standard  tuning fork  of  456  Hz.  The  possible  frequency  of  the  note  B  is (A)  453.5  Hz (B)  455.5  Hz (C)  456.5  Hz (D)  458.5  Hz Solution Ans.  (ABCD) Let  frequency  of  note  B  be  n  then  according  to  question n   =  n–2  or  n  +2 A n   =  n–3  or  n  +3 C As  A  &  C  produce  same  number  of  beats  with  T.F.  of  frequency  456  Hz  so (n–2)  –  456  =  456  –  (n–3)    n  =  458.5  Hz (n+3)  –  456  =  456  –  (n–2)    n  =  455.5  Hz (n+2)  –  456  =  456  –  (n–3)    n  =  456.5  Hz (n+3)  –  456  =  456  –  (n+2)    n  =  453.5  Hz Example#23  to  25 A metallic rod of length 1m has one end free and other end rigidly clamped.  Longitudinal stationary waves are set up in the rod in such a way that there are total six antinodes present along the rod. The amplitude o f   a n  a nt i n o d e   i s   4   ×   1 0–6  m .   Yo u ng ' s   m o d u lu s   a n d   d en s i t y   o f   t he   ro d   are  6. 4   ×   1 0 10  N / m 2  a nd   4   ×   1 0 3  K g / m3 respectively. Consider the free end to be at origin and at t=0 particles at free end are at positive extreme. 2 3 . The equation describing displacements of particles about their mean positions is \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 (A)  s  4  10 6 cos 11 x cos 22  103 t (B)  s  4  10 6 cos  11 x sin 22  103 t  2  2 (C)  s  4  106 cos 5x cos 20  103 t (D)  s  4  106 cos 5x  sin 20  103 t 2 4 . The equation describing stress developed in the rod is (A)  140.8  104 cos  11 x   cos 22  103 t (B)  140.8  104 sin  11 x   cos 22  103 t  2  2 (C)  128  104 cos 5x   cos 20  103 t (D)  128  104 sin 5 x   cos 20  103 t 2 5 . The magnitude of strain at midpoint of the rod at t= 1 sec is (A)  11 3  106 (B)  11 2  10 6 (C)  10 3  106 (D) 10 2  106 E 35

JEE-Physics Solution 23. Ans. (A) 24. Ans.  (B) 25. Ans.  (B) Solution  (23  to  25) Speed of wave  v  y  4  103 5  4                 24 11  /2 Frequency    v  4 103  11 103 Hz ;  Wave Number  K  2  11  4 1 2 11 (i)  Equation of standing wave in the rod  S = A coskx sin(t +) where A = 4 × 10–6 m  at x =0, t =0 S=A A =A cosk(0) sin sin =1 =   2         S  4  106 cos  11 x  cos 22  103 t  2  (ii) Strain = ds  22  106 sin  11 x  co s 2 2   1 0 3 t      str e s s  = Y   × strain dx  2        stress  140.8  104 cos 22  103 t sin  11 x    2 (iii)  Strain at t = 1s  and  x    1 m ;  ds t1  22  106  sin  1 1   11 2  106  4  22 dx  x 2 Example#26  to  28 A detector at x =0 receives waves from three sources each of amplitude A and frequencies f +2, f and f–2. 26. Th e  eq uati o ns   of waves ar e  ; y =A si n[ 2( f+2 )t],  y =A s i n2ft and y =A si n[ 2(f –2)t ].  The t i me at  wh i ch  i nt ensity 2 1 3 is minimum, is \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 (A) t=0, 1/4, 1/2, 3/4,... sec (B) t=1/6, 1/3, 2/3, 5/6, ....sec (C) t=0, 1/2, 3/2, 5/2, ....sec (D) t=1/2, 1/4, 1/6, 1/8, ....sec 2 7 . The time at which intensity is maximum, is (B) t=1/6, 1/3, 2/3, 5/6...sec (A) t=0, 1/4, 1/2, 3/4, ...sec (C) t=0, 1/2, 3/2, 5/2 ...sec (D) t=1/2, 1/4, 1/6, 1/8...sec 2 8 . If I0 A2, then the value of maximum intensity, is (A) 2I (B) 3I (C) 4I (D) 9I 0 0 0 0 36 E

JEE-Physics Solution 26. Ans.  (B) y  y1  y2  y3  A sin 2ft  A sin 2f  2 t  A sin 2 f  2 t  A sin 2ft  2A sin 2ft cos 4t     A 1  2 cos 4 tsin 3ft  A 0 sin 2ft [where A0 = Amplitude of the resultant oscillation = A [1+2cos4t] Intensity   A02   I  1  2 cos 4t 2 For maxima or minima of the intensity. dI  0  2 1  2 cos 4t 2  sin 4t 4  0  1  2 cos 4 t 0  or sin4t =0 dt  cos 4t   1  4t  2n  2   t  n  1  t  1 , 1 , 2 , 5 ... (point of minimum intensity) 2 3 26 6356 27. Ans.  (A) sin 4 t  0  t  n    t  1 13 ... (point of maximum intensity) 0, ,, 4 424 28. Ans.  (D) A t  t = 0 ,  I    ( 1 + 2) 2 A 2= 9 A2     I   =   9I max max 0 Example#29 Consider a large plane diaphragm ‘S’ emitting sound and a detector ‘O’. The diagram shows plane wavefronts for the sound wave travelling in air towards right when source, observer and medium are at rest. AA' and BB' are fixed imaginary planes. Column-I describes about the motion of source, observer or medium and column- II  describes various effects. Match them correctly. AB O S A' B' \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 Column  I Column  II (A) Source starts moving towards right (P) Distance between any two wavefronts (B) Air starts moving towards right will increase. (Q) Distance between any two wavefronts (C) Observer and source both move towards left with same speed. will decrease. (R) The time needed by sound to move from (D) Source and medium (air) both move towards right with same speed. plane AA' to BB' will increase. (S) The time needed by sound to move from plane AA' to BB' will decrease. (T) Frequency received by observer increases. E 37

JEE-Physics Solution A n s .   ( A )    ( Q , T ) ;   ( B )    ( P, S ) ;   ( C )    ( P ) ;   ( D )    ( S , T ) Ve lo c i t y   o f  s o u n d  i n  a   m e di u m  i s   a lway s  g i ven   i n  t he  re fe re n c e  f r a me   o f  m e d i u m . Example#30 Column I represents the standing waves in air columns and string. Column II represents frequency of the note. Match the column-I with column-II. [v = velocity of the sound in the medium] Column  -I Column-II (A) Second harmonic for the tube open at both ends v (P) 4 (B) Fundamental frequency for the tube closed at one end v (Q) 2 (C) First overtone for the tube closed at one end 3v (R) 4 (D)  Fundamental frequency for the string fixed at both ends v (S)  Solution 5v  v (T) 4 Ans. (A) S (B) P (C) R (D) Q For (A) : For open organ pipe 2nd harmonic = 2   2 v For (B) : For closed organ pipe fundamental frequency =  4 3v For (C) : For closed organ pipe, first overtone frequency =  4 v For (D) : For string fixed at both ends, fundamental frequency =  2 Example#31 Two  vibrating  tuning  forks  produce  pr ogressive  waves  given  by  y =  4  sin(500t)  and  y =  2  si n(506t).  These \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 1 2 tuning  forks  are  held  near  the  ear  of  a  person.  The  person  will  hear    beats/s  with  intensity  ratio  between maxima  and  minima  equal  to  .  Find  the  value  of   Solution Ans.  6 y  =  4sin(500  t)                  y  =  2  sin(506  t) 1 2 Number  of  beats   n1  n2  506  500 =  3  beat/sec. 22  As  2 2 2 I    (16)     and   I    4   Imax  I1  I2 2   4  2    6  9 Imin I1  I2  4  2   2  1 2 38 E

JEE-Physics Example#32 A  1000  m  long  rod  of  density  10.0  ×  104  kg/m3  and  having  young's  modulus  Y  =  1011  Pa,  is  clamped at  one  end.    It  is  hammered  at  the  other  free  end  as  shown  in  the  figure.    The  longitudinal  pulse  goes to  right  end,  gets  reflected  and  again  returns  to  the  left  end.    How  much  time  (in  sec)  the  pulse  take  to go  back  to  initial  point? Solution Ans.  2 Velocity  of  longitudinal u Y 1011  103  ms1  10 104  Required  time 2 2 1000   2 s v 103 Example#33 A tuning fork P of unknown frequency gives 7 beats in 2 seconds with another tuning fork Q. When Q is moved towards a wall with a speed of 5 m/s, it gives 5 beats per second for an observer located left to it.  On filing, P gives 6 beats per second with Q. The frequency (in Hz) of P is given by (80 × (I, 0    9) then find the value of  + . Assume speed of sound = 332 m/s. Solution Ans.  9   Q   O  Q  \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 Let f  and f  be the frequencies of tuning forks P and Q,Then | f –f  | = 7/2 1  2  12  v Apparent frequency for O corresponding to signal directly coming from   Q = f    v  v q  2    v   2vqv    vq  Apparent frequency of the echo = f   v    f   =  f    v2  v 2  2  q  2 2 Since, f2 = 5 (g iven)    f2 = 163.5 Hz.   Now, f  = 163.5   3.5  = 167 or 160 Hz, when P is  filed, its frequency 1 will increase, since it is given that filed P gives greater number of beats with Q. It implies that f  must be 167 Hz. 1 E 39

JEE-Physics Example#34 Find  the  number  of  maxima  attend  on  circular  perimeter  as  shown  in  the  figure.  Assume  radius  of circle  >>>. 1.7 S1 S2 Solution Ans.  6  0  1.7 1.7 S1 S2 1.7  0     1  in  each  quadrant,  1  top  point,  1  bottom  point  \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 40 E

JEE-Physics Methods of Work and Energy Work of a Force In everyday life by the word ‘work’, we refer to a vast category of jobs. This meaning is not precise enough to be used as a physical quantity. It was the practical need of scientists and engineers of the late 18th century at the start of Industrial Revolution that made necessary to define work quantitatively as a physical quantity. Physical concept of work involves a force and displacement produced. Work of a constant Force on a body in rectilinear motion To understand concept of work, consider a block being pulled with the help of a string on frictionless horizontal  ground. Let pull F of the string on the box is constant in magnitude as well as direction. The vertical component  F of F , the weight (mg) and the normal reaction N all act on the box in vertical direction but none of them can y move it unless F becomes greater than the weight (mg). Consider that F is smaller than the weight of the box. y y Under this condition, the box moves along the plane only due to the horizontal component F of the force F. x The weight mg, the normal reaction N from the ground and vertical component F all are perpendicular to the y displacement therefore have no contribution in its displacement. Therefore, work is done on the box only by  the horizontal component F of the force F. x  x mg y P x Fy F x  Fx N N Here we must take care of one more point that is the box, which is a rigid body and undergoes translation motion therefore, displacement of every particle of the body including that on which the force is applied are equal. The particle of a body on which force acts is known as point of application of the force.  Now we observer that block is displaced & its speed is increased. And work W of the force F on the block is proportional to the product of its component in the direction of the displacement and the magnitude of the displacement x. W  Fx  x  F cos   x If we chose one unit of work as newton-meter, the constant of proportionality becomes unity and we have node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65  W  F cos   x  F  x   The work W done by the force F is defined as scalar product of the force F and displacement  of point of x application of the force. Unit and Dimensions of Work of a Force SI unit of work is “joule”, named after famous scientist James Prescott Joule. It is abbreviated by letter J. 1 joule = 1 newton × 1 meter CGS unit of work is “erg”. Its name is derived from the Greek ergon, meaning work. 1 erg = 1 dyne × 1 centimeter Dimensions of work are ML2T2 E1

JEE-Physics Example A 10 kg block placed on a rough horizontal floor is being pulled by a constant force 50 N. Coefficient of kinetic friction between the block and the floor is 0.4. Find work done by each individual force acting on the block over displacement of 5 m. F Solution. mg = 100 N Forces acting on the block are its weight (mg = 100 N), normal reaction (N = 100 N) from the ground, force of F = 50 N x = 5 m kinetic friction (f = 40 N) and the applied force (F = 50 N) and displacement of the block are shown in the given F = 40 N figure. N = 100 N All these force are constant force, therefore we use equation Work done W by the gravity i.e. weight of the block       Wif  F  r . mg x g W =0J     Work done W by the normal reaction g x N N W =0J Work done W by the applied force N     F x W = 250 J F Work done W by the force of kinetic friction F f     W = – 200 J f x f Example F A 10 kg block placed on a rough horizontal floor is being pulled by a constant force 100 N acting at angle 37°. Coefficient of kinetic friction 37° between the block and the floor is 0.4. Find work done by each individual force acting on the block over displacement of 5 m. Solution. y Forces acting on the block are its weight mg = 100 N Fy = 60 N (mg = 100 N), normal reaction (N = 40 x N) from the ground, force of kinetic F = 100 N x = 5m node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 friction (f = 16 N) and the applied force 370 (F = 100 N) and displacement of the f = 16 N Fx = 80 N block are shown in the given figure. N = 40 All these force are constant force, therefore we use equation  Work done W by the gravity i.e. weight of the block Wif  F  r . g W =0J      g mg x Work done WN by the normal reaction Work done W by the applied force WN = 0 J     x F N Work done W by the force of kinetic friction  f WF  F  x  Fx x  400 J W = –80 J     f f x 2 E

JEE-Physics Work of a variable Force on a body in rectilinear motion Usually a variable force does not vary appreciably during an infinitely small displacement of its point of application and therefore can be assumed constant in that infinitely small displacement. Fx x x= 0 xi xf x o dx Fx P x= 0 xi x x o x0 f Consider a box being pulled by a variable horizontal force F which is known as function of position x. We now x calculate work done by this force in moving the box from position x to x . Over any infinitely small displacement if dx the force does not vary appreciably and can be assumed constant. Therefore to calculate work done dW by  the force F during infinitely small displacement dx is given by dW  Fx.dx  Fxdx . Integrating dW from x to x x if we obtain work done by the force in moving the box from position x to x . if Wif  xf Fx .dx xi The above equation also suggests that in rectilinear motion work done by a force equals to area under the force-position graph and the position axis. F x xi xf x In the given figure is shown how a force F varies with position coordinate x. Work done by this force in moving x its point of application from position x to x equals to area of the shaded portion. if Example A force which varies with position coordinate x according to equation F = (4x+2) N. Here x is in meters. x Calculate work done by this force in carrying a particle from position x = 1 m to x = 2 m. if Solution.  Using the equation xf 2 4 x  2.dx  8 J Wif  xi Fx .dx , we have Wif  1 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 The above problem can also be solved by using graph Example A horizontal force F is used to pull a box placed on floor. Variation in the force with position coordinate x measured along the floor is shown in the graph. Fx F (N) x(m) 10 x= 0 5 10 15 o x E 3

JEE-Physics (a) Calculate work done by the force in moving the box from x = 0 m to x = 10 m. (b) Calculate work done by the force in moving the box from x = 10 m to x = 15 m. (c) Calculate work done by the force in moving the box from x = 0 m to x = 15 m. Solution. In rectilinear motion work done by a force equals to area under the F (N) B force-position graph and the position axis A 10 (a) W010  Area of trapazium OABC  75 J CE x(m) (b) W1015  Area of triangle CDE  25 J 5 10 15 (c) W015  Area of trapazium OABC  Area of triangle CDE  50 J D Example A coiled spring with one end fixed has a realaxed length l and a spring constant k. What amount of work must 0 be done to stretch the spring by an amount s? Solution. In order to stretch the free end of the spring to a point x, some x x agency must exert a force F, which must everywhere be equal to x= 0 spring force. F = kx The applied force and the spring force are shown in the adjoining kx figure. F The work done WF by the applied force in moving the free end of the spring from x = 0 to x = s be s   F  dx WF  1 ks2 Ans. 2 F 0 0 sx Use of Graph. The variation in F with extension x in the spring is linear therefore area under the force extension graph can easily be calculated. This area equals to the work done by the applied force. The graph showing variation in F with x is shown in the adjoining figure. s  1 ks2 Variation in F with extension x.  F  dx  2 WF Area of the shaded portion  0 Work of a variable Force on a body in curvilinear translation motion node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 Till now we have learnt how to calculate work of a force in y B O Path rectilinear motion. We can extend this idea to calculate work of a  variable force on any curvilinear path. To understand this let us F consider a particle moving from point A to B. There may be  Q dr several forces acting on it but here we show only that force whose  P work we want to calculate. This force may be constant or variable.  Let this force is denoted by F . Consider an infinitely small path length PQ. Over this infinitely small path length, the force can be z x  A assumed constant. Work of this force F over this path length PQ is given by  dW  F  dr 4 E

JEE-Physics The whole path from A to B can be divided in several such infinitely small elements and work done by the force over the whole path from A to B is sum of work done over every such infinitely small element. This we can  calculate by integration. Therefore, work done WAB by the force F is given by the following equation. WA B      rB F dr  rA Work of a Variable Force For a generalization, let point A be the initial point and point B be the final point. Now we can express work   W of a force when its point of application moves from position vector ri to rf over a path by the following if F equation. Wif      rf F dr  ri The integration involved in the above equation must be carried over the path followed. Such kind of integration is known as path integrals. Work of a constant Force  In simple situations where force F is constant, the above equation reduces to a simple form.          rf dr F rf  ri  F  r Wif  F   ri Example  F  Calculate work done by the force 3ˆi  2ˆj  4kˆ N in carrying a particle from point (2 m, 1 m, 3 m) to (3 m, 6 m, 2 m). Solution.   The force F is a constant force, therefore we can use equation Wif  F  r .   W = F . r = (3ˆi  2ˆj  4kˆ ).(5ˆi  5ˆj  5kˆ ) = –15 J Example A particle is shifted form point (0, 0, 1 m) to (1 m, 1 m, 2 m), under simultaneous action of several forces. Two   F1  F2     of the forces are node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 2i  3ˆj  kˆ N and ˆi  2ˆj  2kˆ N . Find work done by these two forces. Solution. Work done by a constant force equals to dot product of the force and displacement vectors.     F r F1  F2  r  W    W Substituting given values, we have    W  3ˆi  ˆj  kˆ  ˆi  ˆj  kˆ  3  1  1  5 J E5

JEE-Physics node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 Work of a force depends on frame of reference A force does not depend on frame of reference and assumes same value in all frame of references, but displacement depends on frame of reference and may assume different values relative to different reference frames. Therefore, work of a force depends on choice of reference frame. For example, consider a man holding a suitcase stands in a lift that is moving up. In the reference frame attached with the lift, the man applies a force equal to weight of the bag but the displacement of the bag is zero, therefore work of this force on the bag is zero. However, in a reference frame attached with the ground the bag has displacement equal to that of the lift and the force applied by the man does a nonzero work. Work and Energy Suppose you have to push a heavy box on a rough horizontal floor. You apply a force on the box it moves and you do work. If you continue pushing, after some time you get tired and become unable to maintain your speed and eventually become unable to push the box further. You take rest and next day you can repeat the experiment and same thing happens. Why you get tired and eventually become unable to pull the box further? The explanation lies in fact that you have a capacity to do work, and when it is used up, you become unable to do work further. Next day you recollect this capacity and repeat the experiment. This capacity of doing work is known as energy. Here it comes from chemical reactions occurring with food in our body and is called chemical energy. Consider another experiment in which we drop a block on a nail as shown in the figure. When set free, weight of the block accelerates it through the distance it falls and when it strikes the nail, its motion vanishes and what appear are the work that drives the nail, heat that increases temperature of the surrounding, and sound that causes air molecules to oscillate. If the block were placed on the nail and pressed hard, it would not have been so effective. Actually, the weight and the distance through which the hammer falls on the nail decide its effectiveness. We can explain these events by assuming that the block possesses energy due to its position at height against gravity. This energy is known as gravitational potential energy. When the block falls, this potential energy is converted into another form that is energy due to motion. This energy is known as kinetic energy. Moreover, when the block strikes the nail this kinetic energy is converted into work driving the nail, increasing temperature and producing sound. Potential, Kinetic and Mechanical Energy If a material-body is moved against a force like gravitational, electrostatic, or spring, a work must be done. In addition, if the force continues to acts even after the displacement, the work done can be recovered in form of energy, if the body is set loose. This recoverable stored energy by virtue of position in a force field is defined as potential energy, a name given by William Rankine. All material bodies have energy due to their motion. This energy is known as kinetic energy, a name given by Lord Kelvin. These two forms of energies - the kinetic energy and the potential energy are directly connected with motion of the body and force acting on the body respectively. They are collectively known as mechanical energy. Other forms of Energy Thermal energy, sound energy, chemical energy, electrical energy and nuclear energy are examples of some other forms of energy. Actually, in very fundamental way every form of energy is either kinetic or potential in nature. Thermal energy which is contribution of kinetic energy of chaotic motion of molecules in a body and potential energy due to intermolecular forces within the body. Sound energy is contribution kinetic energy of oscillating molecules and potential energy due intermolecular forces within the medium in which sound propagates. Chemical energy is contribution of potential energy due inter-atomic forces. Electric energy is 6E

JEE-Physics kinetic energy of moving charge carries in conductors. In addition, nuclear energy is contribution of electrostatic potential energy of nucleons. In fact, every physical phenomenon involves in some way conversion of one form of energy into other. Whenever mechanical energy is converted into other forms or vice versa it always occurs through forces and displacements of their point of applications i.e. work. Therefore, we can say that work is measure of transfer of mechanical energy from one body to other. That is why the unit of energy is usually chosen equal to the unit of work. Work-Kinetic Energy Theorem Consider the situation described in the figure. The body shown is in translation motion on a curvilinear path with increasing speed. The net force acting on the body must have two components – the tangential component necessary to increase the speed and the normal component necessary to change the direction of motion. Applying Newton’s laws of motion in an inertial frame, we have  FT  ma T and FN  ma N FT maT Fn Position 2 man Position 1 Let the body starts at position 1 with speed v and reaches position 2 with speed v . If an infinitely small path  1 2 increment is represented by vector ds , the done by the net force during is work the process     2   2   2 W12  F  ds  FT  FN  ds  1 FTds 1 1 2 v2 1 2 1 m v12 v1 2 2 2 1 maTds   W12  m vdv  m v  The terms 1 m v 2 and 1 m v 2 represent the kinetic energies K and K of the particle at position-1 and 2 2 1 2 2 12 respectively. With this information the above equation reduces to W12  K 2  K 1 The above equation expresses that the work done by all external forces acting on a body in carrying it from one position to another equals to the change in the kinetic energy of the body between these positions. This statement is known as the work kinetic energy theorem. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65How to apply work kinetic energy theorem The work kinetic energy theorem is deduced here for a single body moving relative to an inertial frame, therefore it is recommended at present to use it for a single body in inertial frame. To use work kinetic energy theorem the following steps should be followed. • Identify the initial and final positions as position 1 and 2 and write expressions for kinetic energies, whether known or unknown. • Draw the free body diagram of the body at any intermediate stage between positions 1 and 2. The forces shown will help in deciding their work. Calculate work by each force and add them to obtain work done W by all the forces. 12 • Use the work obtained in step 2 and kinetic energies in step 1 into W12  K 2  K1 . E7

JEE-Physics Example A 5kg ball when falls through a height of 20 m acquires a speed of 10 m/s. Find the work done by air resistance. Solution. The ball starts falling from position 1, where its speed is zero; Position 1 hence, kinetic energy is also zero. K =0J ...(i) 1 During downwards motion of the ball constant gravitational force R Free body diagram of the ball at some mg acts downwards and air resistance R of unknown magnitude h intermediate position mg acts upwards as shown in the free body diagram. The ball reaches position 20 m below the position-1 with a speed v = 10 m/s, so the kinetic energy of the ball at position 2 is K2  1 mv2  250 J ...(ii) 2 ...(iii) The work done by gravity Position 2 v Wg,12  mgh  1000 J Denoting the work done by the air resistance WR,12 and making use of eq. 1, 2, and 3 in work kinetic energy theorem, we have W12  K 2  K 1  Wg,12  WR,12  K 2  K 1  WR,12  750 J Example A box of mass m = 10 kg is projected up an inclined plane from its foot with a speed of 20 m/s as shown in the figure. The coefficient of friction  between the box and the plane is 0.5. Find the distance traveled by the box on the plane before it stops first time. 37° Solution. The box starts from position 1 with speed v = 20 m/s and stops at position 2. 1 Position 2 100 Position 1 37° Kinetic energy at position 1: K1  1 m v 2  2000 J node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 2 1 Kinetic energy at position 2: K = 0 J 2 Work done by external forces as the box moves from position 1 to position 2: W12  Wg,12  Wf,12  60 x  40 x  100 x J Applying work energy theorem for the motion of the box from position 1 to position 2, we have W = k – k  – 100 x = 0 – 2000  x = 20 m 12 2 1 8 E

JEE-Physics Example x =  xo x= 0 A box of mass m is attached to one end of a coiled spring of force x =  xo x= 0 v constant k. The other end of the spring is fixed and the box can slide on a rough horizontal surface, where the coefficient of friction Position 1 Position 2 is . The box is held against the spring force compressing the spring by a distance x . The spring force in this position is more than force mg kx o f = mg of limiting friction. Find the speed of the box when it passes the equilibrium position, when released. N = mg Solution. Before the equilibrium position, when the box passes the position coordinate x, forces acting on it are its weight mg, normal reaction N from the horizontal surface, the force of kinetic friction f, and spring force F = kx as shown in the free body diagram. Let the box passes the equilibrium position with a speed v . o Applying work energy theorem on the box when it moves from position 1 (xo) to position 2 (x = 0), we have W12  K 2  K 1  WF,12  Wf,12  K 2  K 1  1kx2 1 2 vo 2 o 2 o o 2  mgxo  m v 0   k x  2 m gx o Example A block of mass m is suspended from a spring of force constant k. it is held to keep the spring in its relaxed length as shown in the figure. (a) The applied force is decreased gradually so that the block moves downward at negligible speed. How far below the initial position will the block stop? (b) The applied force is removed suddenly. How far below the initial position, will the block come to an instantaneous rest? Solution. (a) As the applied force (F) is decreased gradually, everywhere in its x= 0 downward motion the block remains in the state of translational equilibrium and moves with negligible speed. Its weight (mg) is v1= 0 x0 x kx balanced by the upward spring force (kx) and the applied force. Position-1 v2= 0 mg When the applied force becomes zero the spring force becomes node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 equal to the weight and the block stops below a distance x from Position-2 kx F o the initial position. The initial and final positions and free body diagram of the block at any intermediate position are shown in the adjoining figure. Applying the conditions of equilibrium, we have mg xo  k (b) In the previous situation the applied force was decreased gradually keeping the block everywhere in equilibrium. If the x= 0 applied force is removed suddenly, the block will accelerate v1  0 xm x mg downwards. As the block moves, the increase in spring extension v2 0 increases the upward force, due to which acceleration decreases until extension becomes x . At this extension, the block will acquire o its maximum speed and it will move further downward. When extension becomes more than x spring force becomes more o than the weight (mg) and the block decelerates and ultimately stops at a distance x below the initial position. The initial position- m 1, the final position-2, and the free body diagram of the block at some intermediate position when spring extension is x are shown in the adjoining figure. E9

JEE-Physics Kinetic energy in position-1 is K =0 Kinetic energy in position-2 is 1 K =0 2 Work done W12 by gravity and the spring force is W12  Wg,12  Wspring,12  mgxm  1 k x 2 2 m Using above values in the work energy theorem, we have W12  K2  K1  mgxm  1 k x 2 0 2 m x = 2mg/k m Example A block of mass m = 0.5 kg slides from the point A on a horizontal track with an initial sped of v = 3 m/s 1 towards a weightless horizontal spring of length 1 m and force constant k = 2 N/m. The part AB of the track is frictionless and the part BD has the coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distance AB and BC are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. g = 10 m/s2. v1 AB C D Solution Since portion AB of the track is smooth, the block reaches B with velocity v . Afterward force of kinetic friction 1 starts opposing its motion. As the block passes the point C the spring force also starts opposing its motion in addition to the force of kinetic friction. The work done by these forces decrease the kinetic energy of the block and stop the block momentarily at a distance x after the point C. m v2 = 0 A B C xm D Kinetic energy of the block at position-1 is K1  1 m v 2  2.25 J. Kinetic energy of the block at position-2 is 2 1 Work Wf,12 done by the frictional force before the block stops is K2  1 m v 2 0 J. 2 2 Wf,12  m g(BC  x m )  2.14  x m xm Work Ws,12 done by the spring force before the block stops is Ws,12  kxdx  1 kx 2  x 2 Using above information and the work energy principle, we have 2 m m x 0 W12  K2  K1  2.14  xm  x 2  2.25  xm  0.1 m. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 m The motion of block after it stops momentarily at position-2 depends upon the condition whether the spring force is more than or less than the force of limiting friction. If the spring force in position-2 is more than the force of limiting friction the block will move back and if the spring force in position-2 is less than the force of static friction the block will not move back and stop permanently. Spring force F at position-2 is Fs  kx m  0.2 N. s The force of limiting friction f is fm  smg  1.1 N. m The force of limiting friction is more than the spring force therefore the block will stop at position-2 permanently. The total distance traveled by the block = AB + BC + x =4.24 m. m 10 E

JEE-Physics Conservative and Non-conservative Forces Gravitational, electrostatic, and restoring force of a spring are some of the natural forces with a property in common that work done by them in moving a particle from one point to another depends solely on the locations of the initial and final points and not on the path followed irrespective of pair of points selected. On the other hand, there are forces such as friction, whose work depends on path followed. Accordingly, forces are divided into two categories – one whose work is path independent and other whose work is path dependant. The forces of the former category are known as conservative forces and of the later one as non-conservative forces. A force, whose finite non-zero work W12 expended in moving a particle from a position 1 to another position 2 is independent of the path followed, is defined as a conservative force. Consider a particle moving from position 1 to position 2 along different paths A, B, and C under the action of a conservative force F as shown in figure. If work done by the force along path A, B, and C are W12,A , W12,B and W12,C respectively, we have W12,A  W12,B  W12,C If these works are positive, the work done W21,D by the same force in moving the particle from position 2 to 1 by any other path say D will have the same magnitude but negative sign. Hence, we have W12,A  W12,B  W12,C   W21,D W12,A  W21,D  W12,B  W21,D  W12,C  W21,D  0 The above equations are true for any path between any pair of points-1 and 2. Representing the existing conservative force by  an infinitely path increment by  and integration over a F, dr closed path by     d  , the above equation can be represented in an alternative form as    d   0 r F r The equation () shows that the total work done by a conservative force in moving a particle from position 1 to another position 2 and moving it back to position 1 i.e. around a closed path is zero. It is used as a fundamental property of a conservative force. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 A2 B 1D C • A conservative force must be function only of position not of velocity or time. • All uniform and constant forces are conservative forces. Here the term uniform means same magnitude and direction everywhere in the space and the term constant means same magnitude and direction at all instants of time. • All central forces are conservative forces. A central force at any point acts always towards or away from a fixed point and its magnitude depends on the distance from the fixed point. • All forces, whose magnitude or direction depends on the velocity, are non-conservative. Sliding friction, which acts in opposite direction to that of velocity, and viscous drag of fluid depends in magnitude of velocity and acts in opposite direction to that of velocity are non-conservative. E 11

JEE-Physics Potential Energy Consider a ball of mass m placed on the ground and someone moves it at negligible speed through a height h above the ground as shown in figure. The ball remains in the state of equilibrium therefore the upward force F applied on it everywhere equals to the weight (mg) of the ball. The work Wab done by the applied force on the  ball is Wab  F  h  mgh . F h mg mg v FBD in upward FBD in downward M o tion M o tion C o n fig u ra tio n -a C o n fig u ra tio n -b C o n fig u ra tio n -a Now if the ball is dropped from the height h it starts moving downwards due to its weight and strikes the ground with speed v. The work Wba done by its weight during its downward motion imparts it a kinetic energy K, c which is obtained by using work energy principle and the above equation as Wba  Ka  Kb  mgh  1 m v 2 0  Ka  1 mv2  mgh 2 2 Instead of raising the ball to height h, if it were thrown upwards with a speed v it would have reached the height h and returned to the ground with the same speed. Now if we assume a new form of energy that depends on the separation between the ball and the ground, the above phenomena can be explained. This new form of energy is known as potential energy of the earth-ball system. When ball moves up, irrespective of the path or method how the ball has been moved, potential energy of the earth-ball system increases. This increase equals to work done by applied force F in moving the ball to height h or negative of work done by gravity. When the ball descends, potential energy of the earth ball system decreases; and is recovered as the kinetic energy of the ball when separation vanishes. During descend of the ball gravity does positive work, which equals to decrease in potential energy. Potential energy of the earth ball system is due to gravitation force and therefore is call gravitation potential energy. Change in gravitational potential energy equals to negative of work done by gravitational force. It is denoted by U. In fact, when the ball is released both the ball and the earth move towards each other and acquire momenta of node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 equal magnitude but the mass of the earth is infinitely large as compared to that of the ball, the earth acquires negligible kinetic energy. It is the ball, that acquires almost all the kinetic energy and therefore sometimes the potential energy is erroneously assigned with the ball and called the potential energy of the ball. Nevertheless, it must be kept in mind that the potential energy belongs to the entire system. x= 0 x F kx x= x1 12 E

JEE-Physics As another instance, consider a block of mass m placed on a smooth horizontal plane and connected to one end of a spring of force constant k, whose other end is connected to a fixed support. Initially, when the spring is relaxed, no net force acts on the block and it is in equilibrium at position x = 0. If the block is pushed gradually against the spring force and moves at negligible speed without acceleration, at every position x, the applied force F balances the spring force kx. The work done W by this force in moving the block from 01 position x =0 to x = x is 1 W01x x1   1 k x 2 F  dx 2 1 x 0 If the applied force is removed, the block moves back and reaches its initial position with a kinetic energy K 0, which is obtained by applying work energy theorem together with the above equation. W10  K0  K1  K0  1 mv 2  1 kx12 2 2 x= x1 kx x x= 0 y The above equation shows that the work done on the block by the applied force in moving it from x = 0 to xnode6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 = x is stored in the spring block system as increase in potential energy and when the block returns to its initial 1 position x = 0 this stored potential energy decreases and is recovered as the kinetic energy of the block. The same result would have been obtained if the block were pulled elongating the spring and then released. The change in potential energy of the spring block system when the spring length is increased or decreased by x equals to negative of work done by the spring force. In both the above cases forces involved were conservative. In fact, work done against all conservative forces is recoverable. With every conservative force, we can associate a potential energy, whose change equals to negative of work done by the conservative force. For an infinitely small change in configuration, change in potential energy dU equals to the negative of work done dW by conservative forces. C dW = dU = – dW C Since a force is the interaction between two bodies, on very fundamental level potential energy is defined for every pair of bodies interacting with conservative forces. The potential energy of a system consisting of a large number of bodies thus will be sum of potential energies of all possible pairs of bodies constituting the system. Because only change in potential energy has significance, we can chose potential energy of any configuration as reference value. Gravitational potential energy for uniform gravitational force Near the earth surface for heights small compared to the radius of the earth, the variation in the gravitational force between a body of mass m and the ground can be neglected. For such a system, change in gravitational potential energy in any vertically upward displacement h of mass m is given by U=mgh and in vertical downward displacement h is given by U = – mgh. E 13

JEE-Physics Gravitational potential energy for non-uniform gravitational force r When motion of a body of mass m involves distances from the earth surface The ball at a distance r from large enough, the variation in the gravitational force between the body and the center of the earth. the earth cannot be neglected. For such physical situations the configuration, when the body is at infinitely large distance from the earth center is taken as the reference configuration and potential energy of this configuration is arbitrarily assumed zero (U =0).  If the body is brought at negligible speed to a distance r from infinitely large distance from the earth center, the work done W by the gravitational force is g given by the following equation. Wg r   GMm r Fg  dr  r    Negative of this work done equals to change in potential energy of the system. Denoting potential energies in configuration of separation r and  by Ur and U, we have  GMm Ur  U  Wg  Ur   r Potential energy associated with spring force Relaxed xx The potential energy associated with a spring force of an ideal Compressed x= 0 spring when compressed or elongated by a distance x from its natural length is defined by the following equation Extended U  1 kx2 2 Example Find the gravitational potential energies in the following physical situations. Assume the ground as the reference potential energy level. (a) A thin rod of mass m and length L kept at angle  with one of its end touching the ground. R node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65  (b) A flexible rope of mass m and length L placed on a smooth hemisphere of radius R and one of the ends of the rope is fixed at the top of the hemisphere. Solution In both the above situations, mass is distributed over a range of position coordinates. In such situations calculate potential energy of an infinitely small portion of the body and integrate the expression obtained over the entire range of position coordinates covered by the body. 14 E

JEE-Physics (a) Assume a small portion of length  of the rod at distance  from the h bottom end and height of the midpoint of this portion from the ground  is h. Mass of this portion is m. When  approaches to zero, the gravitational potential energy dU of the assumed portion becomes dU  m ghd  m g sin d LL The gravitational potential energy U of the rod is obtained by carrying integration of the above equation over the entire length of the rod. U L m g sin d  1 mgL sin  0 L 2 (b) The gravitational potential energy dU of a small portion of length  R shown in the adjoining figure, when  approaches to zero is   dU  m gR2 sin d L The gravitational potential energy U of the rope is obtained by carrying integration of the above equation over the entire length of the rope.  L m m   L R L L 1   U gR2 sin d  gR2  cos R    0 Conservation of mechanical energynode6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 The total potential energy of the system and the total kinetic energy of all the constituent bodies together are known as the mechanical energy of the system. If E, K, and U respectively denote the total mechanical energy, total kinetic energy, and the total potential energy of a system in any configuration, we have E=K+U Consider a system on which no external force acts and all the internal forces are conservative. If we apply work-kinetic energy  W12  K 2  K1 theorem, the work W12 will be the work done by internal conservative forces, negative of which equals change in potential energy. Rearranging the kinetic energy and potential energy terms, we have E  K1  U1  K2  U2 The above equation takes the following forms E = K + U = constant E = 0  K + U = 0 Above equations, express the principle of conservation of mechanical energy. If there is no net work done by any external force or any internal non-conservative force, the total mechanical energy of a system is conserved. The principle of conservation of mechanical energy is developed from the work energy principle for systems where change in configuration takes place under internal conservative forces only. Therefore, in physical situations, where external forces or non-conservative internal forces are involved, the use of work energy principle should be preferred. In systems, where external forces or internal nonconservative forces do work, the net work done by these forces becomes equal to change in the mechanical energy of the system. E 15

JEE-Physics Potential energy and the associated conservative force We know how to find potential energy associated with a conservative force. Now we learn how to obtain the conservative force if potential energy function is known. Consider work done dW by a conservative force in moving a particle through an infinitely small path length  as shown in the figure. ds  dU  dW  F  ds  Fds cos  From the above equation, the magnitude F of the conservative force can be expressed. F   dU dU  ds cos  dr If we assume an infinitely small displacement  in the direction of the force, magnitude of the force is given dr by the following equation. F   dU dr Here minus sign suggest that the force acts in the direction of decreasing potential energy. Therefore if we assume unit vector eˆr in the direction of  , force vector  is given by the following equation. dr F    dU eˆr F dr Example Force between the atoms of a diatomic molecule has its origin in the interactions between the electrons and the nuclei present in each atom. This force is conservative and associated potential energy U(r) is, to a good approximation, represented by the Lennard – Jones potential. U(r)  Uo  a 12   a6   r   r    Here r is the distance between the two atoms and U and a are positive constants. Develop expression for the node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 o associated force and find the equilibrium separation between the atoms. Solution. dU Using equation F   , we obtain the expression for the force dr F  6U0 2  a 13   a  7  a   r   r    At equilibrium the force must be zero. Therefore the equilibrium separation r is o 1 ro  2 6 a 16 E

JEE-Physics Potential energy and nature of equilibrium The above equation suggests that on every location where the potential energy function assumes a minimum or a maximum value or in every region where the potential energy function assumes a constant value, the associated conservative force becomes zero and a body under the action of only this conservative force must be in the state of equilibrium. Different status of potential energy function in the state of equilibrium suggests us to define three different types of equilibriums – the stable, unstable and neutral equilibrium. The state of stable and unstable equilibrium is associated with a U point location, where the potential energy function assumes a minimum and maximum value respectively, and the neutral equilibrium is associated with region of space, where the potential energy function assumes a constant value. For the sake of simplicity, consider a one dimensional potential r1 r2 r3 r energy function U of a central force F. Here r is the radial coordinate of a particle. The central force F experienced by the particle equals F to the negative of the slope of the potential energy function. Variation in the force with r is also shown in the figure. At locations r  r1 , r  r2 , and in the region r  r3 , where potential 0 r1 r2 r3 r energy function assumes a minimum, a maximum, and a constant value respectively, the force becomes zero and the particle is in the Force is negative of the slope of the state of equilibrium. potential energy function. Stable Equilibrium. At r = r the potential energy function is a minima and the force on either side acts towards the point r = r . If 11 the particle is displaced on either side and released, the force tries to restore it at r = r . At this location the 1 particle is in the state of stable equilibrium. The dip in the potential energy curve at the location of stable equilibrium is known as potential well. A particle when disturbed from the state of stable within the potential well starts oscillations about the location of stable equilibrium. At the locations of stable equilibrium we have F(r)   U  0 ; and F  0 ; and 2U 0 r r r 2 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 Unstable Equilibrium. At r = r2 the potential energy function is a maxima, the force acts away from the point r = r2. If the particle is displaces slightly on either side, it will not return to the location r = r . At this location, the particle is in the state 2 of unstable equilibrium. At the locations of unstable equilibrium we have F(r)   U  0 therefore F  0 ; and 2U  0 r r r 2 Neutral Equilibrium. In the region r  r, the potential energy function is constant and the force is zero everywhere. In this region, the 3 particle is in the state of neutral equilibrium. At the locations of neutral equilibrium we have F(r)   U  0 therefore F  0 and 2U  0 r r r 2 E 17

JEE-Physics POWER When we purchase a car or jeep we are interested in the horsepower of its engine. We know that usually an engine with large horsepower is most effective in accelerating the automobile. In many cases it is useful to know not just the total amount of work being done, but how fast the work is done. We define power as the rate at which work is being done. Work done Total change in kinetic energy Average Power  Time taken to do work  Total change in time If W is the amount of work done in the time interval t. Then P  W = W2  W1 t t2  t1 When work is measured in joules and t is in seconds, the unit for power is the joule per second, which is called watt. For motors and engines, power is usually measured in horsepower, where horsepower is 1 hp = 746 W. The definition of power is applicable to all types of work like mechanical, electrical, thermal. dW   F.dr Instanteneous power P    F.v dt dt Where v is the instantaneous velocity of the particle and dot product is used as only that component of force will contribute to power which is acting in the direction of instantaneous velocity. • Power is a scalar quantity with dimension M1L2T–3 • SI unit of power is J/s or watt • 1 horsepower = 746 watt Example A vehicle of mass m starts moving such that its speed v varies with distance traveled s according to the law v  k s , where k is a positive constant. Deduce a relation to express the instantaneous power delivered by its engine. Solution Let the particle is moving on a curvilinear path. When it has traveled a distance s, the force F acting on it and its speed v are shown in the adjoining figure. v node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 m F s FN FT Instantaneous power delivered by the engine:     P  F.v  (FT  FN )  v  FT v  ma T v aT dv k2 Tangential acceleration of the vehicle: v  ds 2 From above equations, we have mk3 P s E 2 18

JEE-Physics CIRCULAR MOTION IN VERTICAL PLANE Suppose a particle of mass m is attached to an inextensible light string of length R. The particle is moving in a vertical circle of radius R about a fixed point O. It is imparted a velocity u in horizontal direction at lowest point A. Let v be its velocity at point P of the circle as shown in figure. Here, h = R (1 – cos) ...(i) From conservation of mechanical energy 1 ....(ii) O m(u2 – v2)=mgh  v2 = u2–2gh T 2 P A  mgcos The necessary centripetal force is provided by the resultant of tension mgsin mg T and mg cos T – mg cos = mv2 ...(iii) R Since speed of the particle decreases with height, hence tension is maximum at the bottom, where cos=1 (as =00) mv2 mv 2  T= +mg; T = – mg at the top. Here, v' = speed of the particle at the top. max R min R  Condition of Looping the Loop u  5gR The particle will complete the circle if the string does not slack even at the highest point    . Thus, tension in the string should be greater than or equal to zero (T  0) at =. In critical case substituting T=0 and    in Eq. (iii), we get mg= m v 2 in v = gR (at highest point) m min R PT=0 vmin= gR Substituting    in Eq. (i), Therefore, from Eq. (ii) u2  v2  2gh  gR  2g(2R)  5gR  u = 5gR min min min Thus, if u  5gR , the particle will complete the circle. At u = 5gR , u velocity at highest point is v = gR and tension in the string is zero. A umin= 5gR T=6mg Substituting   00 and v= 5gR in Eq. (iii), we get T =6 mg or in the critical condition tension in the string node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 at lowest position is 6 mg. This is shown in figure. If u < 5gR , following two cases are possible.  Condition of Leaving the Circle 2gR  u  5gR If u < 5gR , the tension in the string will become zero before reaching the highest point. From Eq. (iii), tension in the string becomes zero (T=0) where, cos   v2  cos   2gh  u2 Rg Rg Substituting, this value of cos  in Eq. (i), we get 2gh  u2 h  u2  Rg = h (say) .....(iv) Rg =1 – R h 1 3g E 19

JEE-Physics or we can say that at height h tension in the string becomes zero. Further, if u < 5gR , velocity of the particle 1 becomes zero when 0 = u2 – 2gh  h = u2 =h (say)...(v) i.e., at height h velocity of particle becomes zero. 2 2 2g Now, the particle will leave the circle if tension in the string becomes zero but velocity is not zero. or T = 0 but v 0. This is possible only when h < h 1 2 u2  Rg u2 T= 0  3g < 2g  2u2 + 2Rg < 3u2 u2 > 2Rg u > 2Rg v 0 P O h> R  Therefore, if 2gR < u < 5gR , the particle leaves the circle. R From Eq. (iv), we can see that h > R if u2 > 2gR. Thus, the particle, will leave u A the circle when h > R or 900 <  < 1800. This situation is shown in the figure 2gR < u < 5gR or 900 <  < 1800 Note : After leaving the circle, the particle will follow a parabolic path.  Condition of Oscillation 0 < u  2 g R The particle will oscillate if velocity of the particle becomes zero but tension in the string is not zero or v = 0, but T  0 . This is possible when h <h 21 u2 u2  Rg  2g < 3g  3u2 < 2u2 + 2Rg  u2 < 2Rg  u < 2Rg Moreover, if h = h , u= 2Rg and tension and velocity both becomes zero v= 0 12 T 0 simultaneously. Further, from Eq. (iv), we can see that h  R if u  2Rg . u h R Thus, for 0 < u  2gR , particle oscillates in lower half of the circle (00 <   90°) This situation is shown in the figure. 0  u  2gR or 0°<  90° Example Calculate following for shown situation :– node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 (a) Speed at D (b) Normal reaction at D (c) Height H Solution (a) v2 = v2 – 2gR = 5gR  vD = 5gR D C m v 2 m (5 gR ) (b) mg + N = D ND =  mg = 4mg R DR (c) by energy conservation between point A & C mgH = 1 m v 2  mgR = 1 m v 2  mg2R = 1 m(5gR) + mg2R = 9 mgR  H = 9 R 2 C 2 D 2 22 20 E

JEE-Physics Example A stone of mass 1 kg tied to a light string of length  = 10 m is whirling in a circular path in vertical plane. 3 If the ratio of the maximum to minimum tension in the string is 4, find the speed of the stone at the lowest and highest points Solution Tmax m v 2  mg v 2  g vP Tmin   mg    4   4  v 2  g =4 v p m v 2 p  2 2  4g v 2  5g  4 3v 2 9g  p P P We know v  v   = v 2  g P 10 10  v = 3g = 3 10  = 10 ms–1  v = 7g = 7 10  = 15.2 ms–1 P 3 3  T = (mg + ma) +2m(g+a) (1–cos  ) = m(g+a) (3 – 2cos  ) Example A heavy particle hanging from a fixed point by a light inextensible string of length  , is projected horizontally with speed g . Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle. Solution Let tension in the string becomes equal to the weight of the particle when particle reaches the point B and deflection of the string from vertical is  . Resolving mg along the string and perpendicular to the string, we get net radial force on the particle at B i.e. F = T – mg cos  ....(i) R ....(ii) If v be the speed of the particle at B, then B F= m v 2 O R B T  B A  mgcos From (i) and (ii), we get, T – mg cos  = m v 2 ....(iii) B mgsin mg  node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 Since at B, T = mg  mg 1  cos  = m v 2  v 2  g 1  cos  ...(iv) B B  Applying conservation of mechanical energy of the particle at point A and B, we have 1 mg (1–cos  ) + 1 m v 2 ; where v = g and v = g 1  cos  mv 2 = 2 B A B 2A  g  2g 1  cos  + g (1–cos  ) 2    cos 1  2   cos  = 3  3  Putting the value of cos  in equation (iv), we get : v = g B 3 E 21

JEE-Physics SOME WORKED OUT EXAMPLES Example #1 When a conservative force does positive work on a body, then (A) its potential energy must increase. (B) its potential energy must decrease. (C) its kinetic energy must increase. (D) its total energy must decrease. Solution Ans. (B) Work done by conservative force = – U = positive U Example #2 A box of mass m is initially at rest on a horizontal surface. A constant horizontal force of mg/2 is applied to the box directed to the right. The coefficient of friction of the surface changes with the distance pushed as  = 0x where x is the distance from the initial location. For what distance is the box pushed until it comes to rest again? 2 1 1 1 (A) 0 (B) 0 (C) 20 (D) 40 Solution Ans. (B) Net change in kinetic energy = 0  net work W = 0 N mg x 1 f= N F(mg/2) W   dW   Fdx   Ndx  2 x  mg0 0 xdx  0  x  0 mg Example #3 A car is moving along a hilly road as shown (side view). The coefficient of static friction between the tyres and pavement is constant and the car maintains a steady speed. If, at one of the points shown the driver applies the brakes as hard as possible without making the tyres slip, the magnitude of the frictional force immediately after the brakes are applied will be maximum if the car was at A B (A) point A C (C) point C (D) friction force same for positions A, B and C node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 Solution (B) point B Ans. (C) N N at A & B v2/R at C v2/R mg mg At A & B, N = mg– mv2/R & at C, N = mg + mv2/R  fmax = sN maximum for C 22 E

JEE-Physics Example #4 One end of a light rope is tied directly to the ceiling. A man of mass M initially at rest on the ground starts climbing the rope hand over hand upto a height . From the time he starts at rest on the ground to the time he is hanging at rest at a height , how much work was done on the man by the rope?  (A) 0 (B) Mg (C) – Mg (D) It depends on how fast the man goes up. Solution Ans. (B) Total work done on man = 0 Work done by string = – work done by gravity = –(– Mg) = Mg Example #5 Consider a roller coaster with a circular loop. A roller coaster car starts from rest from the top of a hill which is 5 m higher than the top of the loop. It rolls down the hill and through the loop. What must the radius of the loop be so that the passengers of the car will feel at highest point, as if they have their normal weight? 5m (A) 5 m (B) 10 m (C) 15 m (D) 20 m Solution A Ans. (A) B According to mechanical energy conservation between A and B 5m V GP=0 N Mg mg 5  O  1 mv2  v2  10g ...(i) 2 According to centripetal force equation mv2 for N = mg; 2mg  mv2  r  v2 10g  5m N  mg  r 2g  r 2g node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 Example #6 A pendulum bob of mass m is suspended at rest. A cosntant horiozntal force F = mg/2 starts acting on it. The maximum angular deflection of the string is \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\  (A) 90° (B) 53° mF (D) 60° E (C) 37° 23

JEE-Physics Solution Ans. (C) Let at angular deflection  its velocity be v then by work energy theorem W= KE cos\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ v 1 mv2  mg    cos   F sin   2 sin At maximum angular deflection, v = 0 0 = – mg (1–cos) + mg sin2–2cos = sin 4 + 4 cos2 – 8 cos = sin2 = 1– cos2 2 5 cos2 – 8 cos + 3 = 0 5 cos2 – 5 cos – 3cos + 3 = 0 5cos (cos–1) – 3 (cos–1) = 0 (5cos–3) (cos–1) =0 3 cos = 5 or cos =1  = 37° or  = 0° Example #7 The potential energy for the force   yzˆi  xzˆj  xykˆ , if the zero of the potential energy is to be chosen at the F point (2, 2, 2), is (A) 8 + xyz (B) 8 – xyz (C) 4–xyz (D) 4 + xyz Solution Ans. (B)    U ˆi  U ˆj  U kˆ  U  yz, U  xz, U  xy F x y z x y z Therefore U = – xyz + C where C = constant As at (2, 2, 2), U =0 so C =8 OR Objective question approach : Check that U =0 at (2, 2, 2) Example #8 A particle is projected along the inner surface of a smooth vertical circle of radius R, its velocity at the lowest point 1 being 95Rg . It will leave the circle at an angular distance.... from the highest point 5 (A) 37° (B) 53° (C) 60° (D) 30° Solution Ans. (B) node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 By conservation of mechanical energy [ between point A and B] 1 mu2  mgR 1  cos   1 mv2 N=0 v R B 22  R mg 1  1  2 1 u 2  5  mgR cos  m 95Rg  mgR 1  cos   A 2 95 45 15 3 E   2  2 cos  cos  3 cos    cos       53 25 25 25 5 24

JEE-Physics Example #9 The upper half of an inclined plane with inclination  is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by (A) tan (B) 2tan (C) 2cos (D) 2sin Solution Ans. (B) Refer to figure. In the journey over the upper half of incline, v2 – u2 = 2 as s/2 s s/2 v2 – 0 = 2 (gsin) = gsin.s  2 In the journey over the lower half of incline v2 – u2 = 2 as 0 – gsin.s = 2 g(sin – cos) s –sin = sin – cos   2 sin   2 tan  2 cos  Example #10 Simple pendulums P1 and P2 have lengths 1 = 80 cm and 2 = 100 cm respectively. The bobs are of masses m1 and m2. Initially both are at rest in equilibrium position. If each of the bobs is given a displacement of 2 cm, the work done is W1 and W2 respectively. Then, (A) W1 > W2 if m1 = m2 (B) W1 < W2 if m1 = m2 (C) W1 =W2 if m1 5 (D) W1=W2 if m1 4 m2  m2  4 5 Solution Ans. (A,D) With usual notation, the height through which the bob falls is h   1  cos     2 s in 2   2   2  since  is  2   4  small. Therefore, we can write h  2   a2  a2 . where a = amplitude 2 2    2 mga2 1 Thus, the work done W= P.E. = mgh = 2  W   Example #11 A body of mass m is slowly halved up the rough hill by a force F at which each point is directed along a tangent to the hill. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 F h Work done by the force x (A) independent of shape of trajectory. (C) depends upon h. (B) depends upon x. (D) depends upon coefficient of friction () E 25

JEE-Physics Ans. (ABCD) Solution Work done by the force = Work done against gravity (Wg) + work done against friction (Wf) = Wg   mg sin ds  mg ds sin   mg dh  mgh and Wf   mg cos  ds  mg ds cos   mg dx  mgx F dh  dx Example #12 Ans. (A, C) The kinetic energy of a particle continuously increase with time. Then (A) the magnitude of its linear momentum is increasing continuously. (B) its height above the ground must continuously decrease. (C) the work done by all forces acting on the particle must be positive. (D) the resultant force on the particle must be parallel to the velocity at all times. Solution For (A) : p  2mK if K  then p  For (B) : Its height may  or  For (C) : W = K if K = positive then W = positive For (D) : The resultant force on the particle must be at an angle less than 90° all times Example #13 A particle moves in one dimensional field with total mechanical energy E. If potential energy of particle is U(x), then (A) Particle has zero speed where U(x) = E (B) Particle has zero acceleration where U(x) = E (C) Particle has zero velocity where dU x 0 (D) Particle has zero acceleration where dU x 0 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 dx dx Solution Ans. (A,D) 1 Mechanical energy = kinetic energy + potential energy E = K + U(x) where K = mv2 2 If K = 0 then E = U(x) If F = 0 then F   dU x 0 dU x 0 dx dx 26 E

JEE-Physics Example #14 A spring block system is placed on a rough horizontal surface having coefficient of friction . The spring is given initial elongation 3 m g and the block is released from rest. For the subsequent motion k (A) Initial acceleration of block is 2g. (B) Maximum compression in spring is m g . node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 k k \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ m (C) Minimum compression in spring is zero. \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ m Ans. (A,B,C,D) (D) Maximum speed of the block is 2g k Solution k  3m g   m g  k  For (A) : Initial acceleration =  2g m µmg 2µmg x=0 k k v=0 For (B,C) : Fnet = 0 Therefore maximum compresion = 2mg mg m g and minimum compression = 0   k kk For (D) : At maximum speed Fnet = 0 so by using work energy therorem 1 1  3 m g  2 1  m g  2  2m g  2 2  k  2  k   k  mv2  k  k  m g v = 2g m/k Example #15 to 17 A particle of mass m = 1 kg is moving along y-axis and a single conservative force F(y) acts on it. The potential energy of particle is given by U(y) = (y2–6y+14) J where y is in meters. At y = 3 m the particle has kinetic energy of 15 J. 1 5 . The total mechanical energy of the particle is (A) 15 J (B) 5 J (C) 20 J (D) can't be determined 1 6 . The maximum speed of the particle is (A) 5 m/s (B) 30 m/s (C) 40 m/s (D) 10 m/s 1 7 . The largest value of y (position of particle) is (A) 3+ 5 (B) 3– 5 (C) 3+ 15 (D) 6+ 15 E 27

JEE-Physics Solution 15. Ans. (C) Total mechanical energy = kinetic energy + potential energy = 15+ [32–6(3)+14] = 15 +5 = 20 J 16. Ans. (B) At maximum speed (i.e. maximum kinetic energy), potential energy is minimum U = y2 – 6y + 14 = 5 + (y–3)2 which is minimum at y=3 m so Umin = 5J Therefore Kmax = 20 – 5 = 15 J 1 m v2  15  v max  30 m/s 2 max 17. Ans. (C) For particle K  0 E – U  0 20 – (5+(y–3)2]  0 (y–3)2 15 y–3 15  y  3 + 15 Example #18 to 20 A rigid rod of length  and negligible mass has a ball with mass m attached to one end and its other end fixed, to form a pendulum as shown in figure. The pendulum is inverted, with the rod straight up, and then released. \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\  m 1 8 . At the lowest point of trajectory, what is the ball's speed? (A) 2g (B) 4g (C) 2 2g (D) 8g 1 9 . What is the tension in the rod at the lowest point of trajectory of ball? (A) 6 mg (B) 3 mg (C) 4 mg (D) 5 mg 2 0 . Now, if the pendulum is released from rest from a horizontal position. At what angle from the vertical does the node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 tension in the rod equal to the weight of the ball? (A) cos 1  2  (B) cos 1  1  (C) cos 1  1  (D) c o s 1  1   3   3   2   4  Solution 18. Ans. (B) From COME : 2mg  1 mv2  v  4g  2 g 2 19. Ans. (D) At the lowest point T  mg  mv2  T  mg  m 4g  5mg   28 E

JEE-Physics 20. Ans. (B) mv2  cos Force equation T  mg cos    Energy equation mg cos   1 mv2 T=mg 2 v 1 mg Therefore mg  mg cos   2mg cos   3 cos   1  cos   3 Example #21 AB is a quarter of a smooth horizontal circular track of radius R. A particle P of mass m moves along the track from A to B under the action of following forces : ORB  y F1 = F (always towards y-axis)  F1 F2 x F2 = F (always towards point B) R F3  F3 = F (always along the tangent to path AB) P F4  A F4 =F (always towards x-axis) Column I Column II  (P) 2 FR (A) Work done by F1  1 (B) Work done by F2 (Q) 2 FR (R) FR  (C) Work done by F3  FR (D) Work done by F4 (S) 2 2FR (T)  Solution Ans. (A) (R); (B) (P); (C) (S); (D) (R)  For (A) : Work done by F1  FR node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 For (B) : dW    FR d cos  45     FR  45    d F.ds  2   2   / 4       /2  d F2  2   2  0 R 90° 2 W FR co s 4 5   2 FR sin 45   2FR 0 For (C) : W     F  R   FR F.ds  2  2 For (D) : W     F  R   FR F.ds E 29

JEE-Physics Example #22 A block of mass 2 kg is dragged by a force of 20 N on a smooth horizontal surface. It is observed from three reference frames ground, observer A and observer B. Observer A is moving with constant velocity of 10 m/s and B is moving with constant acceleration of 10 m/s2. The observer B and block starts simultaneously at t =0. BA 10m/s2 10m/s 2kg 20N Column I Column II (A) Work energy theorem is applicable in (P) 100 J (B) Work done on block in 1s as observed by ground is (Q) – 100 J (C) Work done on block is 1 s as observed by observer A is (R) zero (D) Work done on block in 1 s as observed by observer B is (S) only ground & A (T) all frames ground, A & B Solution Ans. (A) (T); (B) (P); (C) (Q); (D) (R) For (A) : Work energy theorem is applicable in all reference frames. For (B) :  20 w.r.t. ground : At t =0, u =0 and t = 1 s, v = at =  2  (1) = 10 m/s For (C) : 11 For (D) : Work done = change in kinetic energy = (2) (10)2 – (2) (0)2 = 100 J 22 w.r.t. observer A : Initial velocity = 0 – 10 = – 10 m/s, Final velocity = 10 – 10 = 0 11 Work done = (2) (0)2 – (2) (–10)2 = – 100 J 22 w.r.t. observer B : Initial velocity = 0 – 0 = 0 Final velocity = 10 – 10 = 0; Work done = 0 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Work Power Energy\\Eng\\Theory.p65 30 E