OSCILLATION - Lecture Notes

NEET 2022 [LECTURE NOTE - PCB] PHYSICS OSCILLATIONS Periodic Motion OR Harmonic Motion Any motion that repeats itself at regular intervals of time is known as periodic motion Eg. Revolution of earth around the sun Oscillatory Motion A body is said to be in oscillatory motion if it undergoes a to and fro motion about an equilibrium position or mean position Eg. Motion of a simple pendulum * Every oscillatory motion is periodic, but every periodic motion need not be oscillatory  Example, circular motion is periodic, but not oscillatory * Time period (T) * Time for one oscillation * SI unit is second(s) * Frequency  No. of oscillations per second  1 T  SI unit (Hz) or s–1 1Hz = 1s–1 There is no significant difference between oscillations & vibrations. If frequency is high, we call it as vibration. If frequency is low, it is oscillation Displacement In oscillatory motion, the term displacement means any physical quantity that changes with time. Eg. In the case of oscillation of heart, displacement means volume In oscillatory motion, displacement can be represented by using periodic functions. A function f(t) is said to be periodic if and only if 5

Brilliant STUDY CENTRE f t  f T  t  where T is the period of function eg: sin   sin   2 & cos   cos   2 Simple Harmonic Motion [SHM] A motion is said to be simple harmonic, if its acceleration at any instant is directly proportional to displacement from mean position, and acceleration is always directed towards the mean position. i.e., a  - displacement [Displacement is always away from mean position & it is always measured from mean position] a - y a = 2 y a  2y  0 i.e., d2y  2 y  0 dt 2 This is the equation of a SHM. Solution of this equation gives displacement of SHM, which is given by y  A sin t   OR y  A cos t   where A  Amplitude : Maximum displacement   Angular velocity OR Angular frequency   2  2 T  SI unit of  is rad/s t    phase   initial phase OR phase constant OR epoch 6

NEET 2022 [LECTURE NOTE - PCB] If   0 y  A sin t y  A cos t when t  0 when t  0 y0 yA  y  A sin t , represents oscillation  y  A cos t , represents, oscillation starting from mean position starting from extreme position  y  A sin t   represents particle  y  A cos t  . represents particle initially shifted from mean position initially shifted from extreme position by an angle  by an angle  Oscillation of a particle along a straight line Draw the displacement - time graph of above motion 7

Brilliant STUDY CENTRE * Velocity in SHM Let y  A sin t   v  dy  Acost   dt v  A 1 sin2 t   v   A2  A2 sin2 t   v   A2  y2  Relation between velocity & displacement in SHM  At mean position (y = 0)  vmax  A  At extreme position ( y = A) v0 * Acceleration in SHM We have y  A sin t   v  A cos t   a  2A sin t   i.e., a  2y At mean position (y = 0) a0 At extreme position (y = A) amax  2A 8

NEET 2022 [LECTURE NOTE - PCB] Comparison of displacement-time graph, velocity-time graph, acceleration-time graph Displacement - Velocity graph v   A2  y2 v2  A2  y2 2 y2  v2  A2 2 y2  v2 1 Equation of ellipse A2 2 A 2 9

Brilliant STUDY CENTRE Acceleration - displacement graph Force in SHM Let ‘m’ be the mass of a particle in SHM. Then force acting on it is F = ma where a  2y F  m2y where m2  k force constant F  ky –ve sign indicates that force is restoring Here also, force maximum at extreme position Fmax  KA 10

NEET 2022 [LECTURE NOTE - PCB] Energy in SHM 1) Kinetic energy KE  1 mv2 2 where v   A2  y2 KE  1 m2  A 2  y2  2 KE  1 k A2  y2  2 At mean position (y = 0)  kmax  1 kA2 2 At extreme position (y = A) KE = 0 2) Potential Energy Potential energy of an oscillating particle at a point is defined as work done by an external agent to bring the body from mean position to that point Let ‘y’ be the displacement from mean position, then potential energy is given by U  1 ky2 2 At mean position (y = 0)  PE = 0 At extreme position (y = A) PEmax  1 KA2 2 Total Energy E = KE + PE E  1 m2 A2  y2   1 m2A2 2 2 11

Brilliant STUDY CENTRE E  1 m2A2 2 E  1 kA2 Total Energy of the oscillating particle always remains a constant 2 Relation between SHM & uniform circular motion  Consider a uniform circular motion A  Radius of circular path B  Angular velocity of circular motion Let ‘p’ be the position of the particle at t = 0, and p/ be the position at time t. 12

NEET 2022 [LECTURE NOTE - PCB] Projection of position vector op/ on the diameter CD is y  A sin t This is displacement equation of SHM From the figure it is clear that particle only execute circular motion. But foot of the perpendicular drawn from the particle to the diameter CD will execute a SHM. Therefore SHM can also be defined as the projection of a uniform circular motion, on any of the diameter. NOTE 1 : 1) Time taken to displace a particle from 0 to A is T 2 12 2) Time taken to displace the particle from A to A is T 2 6 Proof 0A2 y  A sin t1 13

Brilliant STUDY CENTRE A  A sin t1  sin t1  1 2 2 t1   6 t1    T  T 6 6 2 12 A A 2 t2  T  t1  T T  3T  T  T 4 4 12 12 6 NOTE : 2 If two SHM are represented by y1  A1 sin 1t  1  & y2  A2 sin 2t  2  Then phase difference between them is   2t  2   1t  1   If  is +ve, y2 leads y1  If  –ve, y1 leads y2  If   0, 2, 4,........  Then they are said to be in  phase If   ,3,5,......  Then they are said to be out of phase NOTE : 3 If a SHM is represented by y  A1 sin t  A2 sin t   Then it can be represented as y  A sin t   where resulted amplitude 14

NEET 2022 [LECTURE NOTE - PCB] A A12  A 2  2A1A 2 cos  2 where tan   Bsin  A  B cos  Applications of SHM 1) Oscillations of a spring Consider a light and elastic spring suspended vertically from a rigid support. Let a body of mass ‘m’ be attached to the lower end of the spring   elongation or extension From figure (B), according to Hooke’s Law, restoring force is directly proportional to elongation i.e. F   OR F  k | k  spring constant –ve sign indicates that force is restoring At equilibrium mg  k  k  mg  15

Brilliant STUDY CENTRE Now the body be pulled further down through a small distance ‘y’ and released, it starts vertical oscillations. From fig. (c) Fnet  mg  k   y ma  mg  k  ky i.e. ma = –ky a    k  y  m  This equation of the form a  2y .  Motion of spring is SHM By comparing   k m & k  m2  Time period of oscillation of a spring-block system is T  2  2 m k  Spring constant ‘k’ is independent of acceleration due to gravity g NOTE 1 : If the spring is not light, but has a mass m5, then time period of oscillation of spring block system is m  1  ms 3 T  2 k NOTE 2 : T  2 m k T  2 m k 16

NEET 2022 [LECTURE NOTE - PCB] NOTE : 3 NOTE : 4 T  2 m1  m2  k  Where restoring force is less than frictional force between m1 & m2 Combination of Springs a) Parallel combination  For parallel combination both springs have same elongation. Let ‘  ’ be the elongation, then mg  k1  k2 = k1  k2    i.e., mg  kp  where kp  k1  k2  Effective spring constant in parallel combination 17

Brilliant STUDY CENTRE  T  2 m kp b) Series combination In series combination force is same in both springs, but elongation is different For first spring k11  F . 1  F k1 For second spring k  2 F 2   F 2 k2  Let ‘  ’ be the total elongation of springs, then     2 1 F FF  ks k1 k2 18

NEET 2022 [LECTURE NOTE - PCB] i.e., 1 11 ks k1 k 2 where ks  effective spring constant in series combination i.e., ks  k1k 2 k1  k2 NOTE : 1  Spring cons tan t  1 length of spring NOTE : 2 If a spring of spring constant ‘k’ is cut into ‘n’ equal parts, then spring constant of each part becomes nk. 2) Oscillations of a simple pendulum Simple pendulum consist of a light & inextensible spring connected to a small mass or bob m  mass of bob   length of pendulum (Distance between point of suspension & centre of mass of bob) From figure (B) T  mg cos   FC  centripetal force T  mg cos   mv2  19

Brilliant STUDY CENTRE Restoring torque is provided by mg sin  i.e., restoring torque    mg sin sin 90  ve sign indicates restoring i.e., I  mg sin  torque for small oscillations sin    I  mg     mg   I  moment of inertia  I  This is the form of a  2y where I  m2 It is a simple harmonic motion. By comparing   mg I mg g    m2   Time period of oscillation T  2  2  g 1) Time period is independent of mass of bob 2) T   3) T  1 g Graph 1)   T2 T2  42  Slope  42 T2   g g 20

NEET 2022 [LECTURE NOTE - PCB] 2)   T graph T2  42   Eqn. of parabola g  y2  4ax    T2 &   T graph intersect at T = 1 NOTE Second’s pendulum : A pendulum whose time period is 2 second * Length of second’s pendulum on the surface of earth is approximately 1m Special Cases * If a pendulum is suspended from the roof of a lift moving vertically upwards with an acceleration ‘a’, then effective acceleration on the bob becomes (g + a) Time period of oscillation 21

Brilliant STUDY CENTRE  T  2 ga * If a pendulum is suspended from the roof of a lift moving vertically downwards with an acceleration ‘a’, then effective acceleration on the bob becomes (g – a)  T  2 ga * If a pendulum is suspended on the roof of a car moving horizontally with an acceleration ‘a’, then effective acceleration on the bob becomes geff  g2  a 2 22

NEET 2022 [LECTURE NOTE - PCB]  In such a case mean position of the pendulum is shifted backwards by an angle tan   a g * If a pendulum is on the roof of a car moving down an inclined plane of inclination ‘  ’, then effective acceleration on the bob becomes g cos    T  2 g cos  * 23

Brilliant STUDY CENTRE Net force on the bob is F = mg – FB mgeff = mg – weight of liquid displaced V  volume of bob V geff  V g  V g geff  g  g geff       g  g eff      g    1       T  2  1    g   * Torsional Pendulum  It consists of a horizontal circular disc suspended with the help of wire  If it is twisted by an angle  & then released, it executes torsional oscillations Restoring torque   C C  restoring torque per unit twist I  C    C   a  2y I SHM By comparing   C I  moment of inertia of disc I  T  2 I C 24

NEET 2022 [LECTURE NOTE - PCB] U-Tube Oscillator A  Area of cross section h  height of liquid in the tube in the state of equilibrium d  density of liquid If we withdraw applied force, then liquid will oscillate  Here restoring force is provided by the weight of liquid in ‘2x’ height. i.e., F = –weight of liquid in 2x height = A 2x dg M  Total mass of liquid in the tube Ma  2Adg x a    2Adg  x  a  2y  M  SHM By comparing   2Adg M  T  2 M 2Adg But M  A2hd T  2 A2h.d 2Adg T  2 h g 25

Brilliant STUDY CENTRE Damped Oscillations Oscillations in which amplitude decreases gradually with the passage of time are called damped oscillations. The restoring forces are known as damping forces. Generally damping force is directly proportional to velocity of the oscillating body. Fd  V  Fd  bV b  Damping constant, it depends on size and shape of the body and nature of the medium  –ve sign indicates that damping force is opposite to the direction of motion Net force on the body F = restoring force + damping force ma = –ky – bv ma + bv + ky = 0 md2 y  bdy  ky  0 dt 2 dt  This is the equation of damped oscillator. Solution of this equation gives displacement of damped oscillator, which is given by, y   bt cos 1t   2m Ae where  bt Amplitude  Ae 2m A  initial amplitude 1  Angular velocity of damped oscillator 1  k  b2 m 4m2 26

NEET 2022 [LECTURE NOTE - PCB] If b2  km [small damping] 1    k m Total Energy of the oscillator E  1K Amp2 2 E  1 k.A2e  bt m 2 i.e.  bt E  E0e m E0  1 kA 2 initial energy 2 Total Energy also decrease with time Forced Oscillation A body oscillates under the influence of an external periodic force is known as forced oscillation Let F(t) be the external periodic force Ft  F0 cos dt F0  Amplitude of applied force d  Angular velocity of applied force OR driving force  Net force on the body F  F t   Fres  Fdam ma  F0 cos dd  ky  bv 27

Brilliant STUDY CENTRE ma  bv  ky  F0 cos dt m d2y  b dy  ky  F0 cos d t dt 2 dt  This is the equation of a forced oscillator. Solution of this equation gives displacement of forced oscillator, which is given by y  A cos dt   A F0 1    where Amplitude m2 2  d2 2  2db2 2 Initially the oscillator oscillates with its natural frequency  . When we apply an external periodic force, oscillations with natural frequency die out & body will oscillates with the frequency of force d  Resonance If the frequency of applied force is equal to natural frequency of the body then the body will oscillates with maximum amplitude. This is known as resonance. At resonance   d  Amplitude A max  F0 db NOTE :  In ideal case of zero damping amplitude of the oscillator at resonance is infinity. 28