Comprehensive CBSE Question Bank In Physics-XI (Term-II)

302 Physics—XI Time period,  or  Also, mg = kl or   remains Any change in g will bring about a corresponding change in l so that constant. So, T is independent of g. Frequency,      12. A 5 kg collar is attached to a spring of spring constant 500 N m–1. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate (a) the period of oscillation, (b) the maximum speed and (c) maximum acceleration of the collar. [NCERT Solved Example] Ans. (a) Period of oscillations is given by T=     = (2/10) s = 0.628 s (b) Maximum speed, vmax. =    =   = 1 m s–1 (c) The acceleration of the collar at the displacement y from the equilibrium is given by, a(t) =     Therefore the maximum acceleration is, amax. =     = 10 m s–2 13. Two identical springs, each of spring factor k, may be connected in different ways as shown in Fig. 6.34 (a, b, c). Deduce the spring factor of the oscillation of the body in each case. Also find the frequency in each case. Fig. 6.34 Ans. Each spring has a spring factor k. So, for an extension y in the spring, the restoring force is given by F = –ky

Oscillations 303 Case (a) [Fig. 6.34 (a)]. Let the load P be pulled down through a small distance y. Both the springs will suffer the same extension y. Total restoring force = 2F If k1 be the spring factor of the combination, then 2F = – k1y or  Comparing with F = – ky, we get = k or k1 = 2 k =  Case (b) [Fig. 6.34 (b)]. Let each spring be made to suffer extension y. Let k2 be the spring factor of the combination. Restoring force, F = –k2 (2y) Comparing with F = –ky, we get 2k2 = k or  = =   Case (c) [Fig. 6.34 (c)]. Let the load P be pulled down through a small distance y. The upper spring will be extended while the lower spring will be compressed. Each spring will give rise to a restoring force F. Total restoring force = 2F If k3 be the spring factor of the combination, then 2F = – k3 y or  Comparing with F = – ky, we get = k or k3 = 2k  =  14. Two rigid bodies A and B of masses 1 kg and 2 kg respectively are A rigidly connected to a spring of force constant 400 N m–1. The body B rests on a horizontal table (Fig. 6.35). From the rest position, the body A is compressed by 2 cm and then released. Deduce (i) the frequency of oscillation, (ii) total oscillation energy, (iii) the ampiltude of the harmonic vibration of the reaction of the table on body B. Ans. Since it is the body A which is oscillating,  m = 1 kg (i) frequency of oscillation,  = B  or  = 3.182 Hz Fig. 6.35   (ii) Total oscillation energy, E =  or E =      =  = 0.08 J [... A = 2 cm = 0.02 m]

304 Physics—XI (iii) Total force acting on the table = (1 + 2) kg wt. = 3 kg wt. = 3 × 9.8 N = 29.4 N This is the mean upward reaction. Now, due to oscillations, the maximum tension developed in the spring is given by F = kA = 400 × 0.02 N = 8 N Thus, the amplitude of the vibration of reaction on the table is 8 N. In other words, the net reaction of the table will vary from (29.4 + 8) N to (29.4 – 8) N, i.e., between 37.4 N and 21.4 N. 15. Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in Fig. 6.36. Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations. [NCERT Solved Example] m kk Fig. 6.36 Ans. Let the mass be displaced by a small distance x to the right side of the equilibrium position, as shown in Fig. 6.37. Under this situation, the spring on the left side gets elongated by a length equal to x and that on the right side gets compressed by the same length. The forces acting on the mass are then, F1 F2 O Fig. 6.37 F1 = – kx (force exerted by the spring on the left side, trying to pull the mass towards the mean position) F2 = – kx (force exerted by the spring on the right side, trying to push the mass towards the mean position) The net force, F, acting on the mass is then given by, F = – 2kx. Hence the force acting on the mass is proportional to the displacement and is directed towards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is, T =  . 16. A block of mass m is placed on a frictionless surface and is connected to two springs of force constants k1 and k2 as shown in Fig. 6.38. Calculate the time period of oscillation of the block.

Oscillations 305 k1 k2 m Fig. 6.38 Ans. Suppose block is displaced to the left through a small distance x. Now, the left spring shall get compressed and the right spring shall be elongated. Let F1 and F2 be the restoring forces in the springs of force constants k1 and k2. Then, F1 = – k1x and F2 = – k2x Total restoring force, F = F1 + F2 or F = – k1x – k2x or F = – (k1 + k2)x ...(1) If k be the force constant of the system, then F = – kx ...(2) Comparing (2) and (1), k = k1 + k2 Time period, T=  or  .  17. What is a simple pendulum? Derive an expression for the time period of a simple pendulum. Ans. An ideal simple pendulum consists of a heavy point mass suspended by a massless, inextensible and perfectly flexible string from a rigid support of infinite mass about which it is free to oscillate. Fig. 6.39 shows a simple pendulum of length L S suspended from a rigid support S. The bob of mass m is free to oscillate in a plane about the vertical  line through the support. T A SO is the equilibrium or mean or the undisplaced position of the simple pendulum. SA represents the mg sin   mg cos  displaced position of the pendulum at any time such mg that ASO =  (angular displacement). When the O bob is at mean position,  = 0°. Fig. 6.39. Vibrating simple There are only two forces acting on the bob : pendulum (i) Weight mg of the bob acting vertically downwards through the centre of gravity of the bob. Here, g is the value of acceleration due to gravity at the place where the pendulum is suspended. (ii) Tension T in the string acting along the length of the string and directed towards the point of suspension S. The force mg can be resolved into two rectangular components—mg cos  along the string and mg sin  perpendicular to it. The motion of the bob is along a circular arc of radius L. The centre of the arc is at the support point. The bob has a radial acceleration 2L and also a tangential

306 Physics—XI acceleration. The tangential acceleration is due to the fact that the motion along the arc of the circle is not uniform. The radial acceleration is provided by the net radial force T – mg cos . The tangential acceleration is provided by mg sin . The radial force gives zero torque about the support. Torque  about the support is entirely provided by the tangential component mg sin  of the weight mg.   = – (mg sin ) L ...(1) This is a restoring torque which tends to reduce angular displacement. This explains negative sign in the above equation. By Newton’s law of rotational motion,  = I ...(2) where I is the moment of inertia of the system about the support and  is the angular acceleration. Comparing (2) and (1), I = – mg L sin or  = – mgL sin  I We know that sin  =  –  +  – ......, where  is in radian. If  is small, then sin  can be approximated by .  =–  ...(3) This equation tells us that the angular acceleration of the pendulum is proportional to the angular displacement  but opposite in sign. So, we may conclude that the motion of a simple pendulum swinging through only small angles is approximately SHM. Note that the angular amplitude m of the motion (the maximum angle of swing) must be small. Equation (3) is mathematically identical to the equation  = – 2.  or T = 2  = ; = But I = mL2  T = 2 or T = 2 Frequency,     The time period T is independent of the mass m of the bob and the amplitude of its oscillation. Such vibrations where time period is independent of amplitude are called isochronous vibrations. At a given place ( g will be fixed), it depends only on the length L of the pendulum. 18. The bottom of a dip on a road has a radius of curvature R. A rickshaw of mass M left a little away from the bottom oscillates about the dip. Deduce an expression for the period of oscillation. Ans. When the rickshaw is away from O (Fig. 6.40), the following forces act on it : (i) Weight Mg of the rickshaw acting vertically downwards through its centre of gravity.

Oscillations 307 (ii) Normal reaction offered by the road to the rickshaw. The weight Mg can be resolved into two rectangular  components—Mg cos  and Mg sin . The RN component Mg cos  is balanced by reaction. The only unbalanced force is Mg sin .  Magnitude of restoring force, F = Mg sin  Spring factor, k=   Mg A Since  is very small    sin   Mg cos   O Mg   k=  Fig. 6.40 Time period, T =    = 19. Show that if a liguid initially at rest in U-tube is disturbed, it will undergo SHM. Find the expression for the time period of these Oscillations. Ans. Consider a liquid column abcd of length L in a U-tube of uniform cross-sectional area A. Fig. 6.41 (i) shows the static equilibrium position of the liquid column. Let the liquid level in the left limb be depressed through a distance y. The liquid level in the right limb will go up by the same distance y [Fig. 6.41 (ii)]. The external force applied on the liquid level in the left limb will be balanced by an equal and opposite restoring force. The restoring force will be equal to the weight of liquid column of length ‘2y’. As soon as the external force ceases to act, the liquid column in the U-tube will begin to oscillate up and down about its mean position. y ad 2y y bc Fig. 6.41. Oscillations of a liquid column Restoring force, F = – weight of height ‘2y’ of liquid column or F = – (2y × A) g = – (2A g)y where  is the density of the liquid and ‘g’ is the value of acceleration due to gravity. Now, F=–ky where k( = 2Ag) is a force constant of the system. Mass of oscillating liquid column, m = (L A)  Acceleration, =  or =  or = 

308 Physics—XI Put = 2.  = –2 y The acceleration is directly proportional to displacement and is directed towards the mean position. Thus, the motion of the oscillating liquid is simple harmonic motion. Time period, T =  or T=     or T =   Frequency,  =   The time period T is independent of the mass of the liquid column, density of liquid and the cross-sectional area of the U-tube. However, it depends upon the length of the liquid column and also on the value of acceleration due to gravity. 20. A mercury column of mass m oscillates in a U-tube. One centimetre of the mercury column weighs 15 g. Calculate (a) spring constant of motion and (b) period of oscillation. Ans. Let the liquid in the right arm be depressed through a distance y metre. Then the liquid in the left arm will rise through a distance y metre. Restoring force, F = – weight of unbalanced column of length 2y metre or F=     y  y or Comparing with F = – 29.4 y F = – ky, we get k = 29.4 N m–1 (b) T =    Fig. 6.42 = = 1.159 21. Derive an expression for the time period of a floating cylinder. Ans. Consider a cylinder of mass m, length L and uniform area of cross- section A. m = AL  where is the density of material of the cylinder. Suppose the cylinder is floating vertically in a liquid of density. Let l be the length of the cylinder dipping in the liquid in equilibrium position [Fig. 6.43]. According to Archimedes’ principle, the upward thrust F1 acting on the cylinder is equal to the weight of the liquid displaced by length l of the cylinder.

Oscillations 309 Fig. 6.43. Floating cylinder F1 = Al  g Since the cylinder is in equilibrium position,  mg = A l g or m = Al  Let the cylinder be pushed down through a small distance y. Now, the upward thrust F2 on the cylinder is equal to the weight of the liquid displaced by length ( l + y) of the cylinder. F2 = A(l + y)  g Restoring force, F = – (F2 – mg) = – [A(l + y) g – Al  g] or F = – (A  g) y F y F is directed towards equilibrium position. So, the motion of cylinder is SHM. Now, F = – ky  k = A  g, where k is spring factor Inertia factor, m = AL  T = 2 inertia factor  2 AL  2 L spring factor Ag g Using m = A l , T = 2 Al   2 l A g g Frequency,  = 1  1 g C T 2 l y 22. A weighted glass tube is floating in a liquid with 20 cm B of its length immersed. It is pushed down a little and l released. Calculate the time period of its vibration. Fig. 6.44 Ans. Let B be the point up to which the weighted glass tube is immersed in the liquid (Fig. 6..44). l = 20 cm = 0.20 m Applying the law of floatation, weight of tube = weight of liquid displaced = upthrust

310 Physics—XI  mg = (Al) g where m is the mass of the tube, A is the cross-sectional area of the tube and  is the density of liquid. Now, m = Al = A × 0.20 ×  Let the tube be pushed into the liquid up to the point C such that BC = y Restoring force = weight of extra liquid displaced = additional upthrust = – Ayg = – (Ag)y or – ky = – (Ag)y  k = Ag Time period, T =        or T =  s = 0.898 s. x O 23. For the damped oscillator shown in Fig. 6.45 Rigid support the mass m of the block is 200 g, k = 90 N m–1 Spring and the damping constant b is 40 g s–1. Calculate (a) the period of oscillation, (b) time Block taken for its amplitude of vibrations to drop to half of its initial value and (c) the time taken Surrounding for its mechanical energy to drop to half its medium initial value. [NCERT Solved Example] Fig. 6.45 Ans. (a) k m = 90 N m–1 × 200 × 10–3 kg = 18 N m–1 kg = 18 kg2 s–2 = 4.243 kg s–1 b = 40 × 10–3 kg s–1 = 0.040 kg s–1 Clearly, b << , << 1 So, the damping is small. In the equation  =  , is negligibly small.   = . So, we use the formula T = 2 . T = 2 s = 0.296 s = 0.3 s (b) Amplitude = A e–bt/2m, = A e–bT1/2/2m or ebT1/2/2m = 2 or = loge 2 or T1/2 = =  s = 6.93 s

Oscillations 311 (c) E(t) = kA2 e–bt/m, E(t) = Ei e–bt/m, where Ei    is the initial energy. = Ei e–bT1/2/m   or = loge 2 or T1/2 = =  s = 3.465 s This is just half of the decay period for amplitude. This should not be surprising because energy depends upon the square of amplitude. 24. A simple pendulum in a stationary lift has time period T. What would be the effect on the time period when the lift (i) moves up with uniform velocity v (ii) moves down with uniform velocity v (iii) moves up with uniform acceleration a (iv) moves down with uniform acceleration a (v) begins to fall freely under gravity? Ans. (i) and (ii). Since acceleration of the lift is zero therefore there will be no effect on time period. (iii) When the lift moves up with uniform acceleration a, the effective value of acceleration due to gravity is g + a.      (iv) When the lift moves down with uniform acceleration a, then the effective value of g is g – a.     (v) When the lift begins to fall freely under gravity, the effective value of g becomes zero. So, T is infinite i.e., the simple pendulum shall not oscillate. 25. Two springs have their force constants k1 and k2 (k1 > k2 ). Determine the spring on which more work is done when (i) their lengths are increased by the same amount, and (ii) they are stretched by the same force. Ans. (i) Let each spring be stretched through a distance y. Work done on second spring = Work done on first spring = But k1 > k2 (given)   So, more work is done on the first spring. (ii) Let each spring be stretched by the same force F. Let y1 and y2 be the extensions in the first and second springs respectively. Then, y1 = 

312 Physics—XI Work done on first spring =      Similarly, the work done on the second spring = But  (given)   So, more work is done on the second spring. 26. A vertical U-tube of uniform cross-section contains water y y up to a height of 0.3 m. Show that if water on one side is 2y depressed and then released, its motion up and down the two sides of the tube is simple harmonic motion. 0.3 m Calculate its time period. Fig. 6.46 Ans. Length of water column = 2 × 0.3 metre = 0.6 metre mass of water, m = 0.6 × A × 1000 kg = 600 A kg where A m2 is the cross-sectional area of the U-tube and 1000 kg m–3 is the density of water. Restoring force, F = – (weight of water column of length 2y) = – (2y × A × 1000 × 9.8) N Acceleration, =     or =  So, the acceleration is directly proportional to displacement and is directed towards the mean position. Hence, the motion of the water column is simple harmonic. Time period, T =      = 1.1 s. 27. Fig. 6.47 shows a solid cylinder of mass m and radius r attached to a horizontal massless spring of spring constant k. The cylinder can roll without slipping. When the cylinder is pulled and released, its centre of mass executes simple harmonic motion. Calculate the periodic time. r Fig. 6.47 Ans. Total energy, E = kx2 + mv2 + I2 or E = kx2 + mv2 +   or E = kx2 + mv2   In SHM, total energy is conserved.

Oscillations 313  = 0  k(2x) + m(2v) = 0 m = – kx or = – x, 2 = ,  = So, T = 2 . 28. Prove that the time period of a simple pendulum of infinite length is not infinite but 84.6 minutes. l Ans. Time period of simple pendulum, T = 2 g . From this equation, it is clear that T will become infinite T () if the length of the pendulum is infinite. But this is cos  not the case. In-fact, time period of a simple pendulum O O of infinite length is 84.6 minutes and not infinite.  A simple pendulum of infinite length can be considered R sin  to oscillate along a straight line OO. Let OO = x. EARTH At the displaced position O, the weight (mg) of the pendulum acts towards the centre of the Earth. Let weight (mg) of the pendulum make an angle  with the horizontal line OO. Resolve mg into two components : The vertical component of mg sin  is balanced by tension (T) in the string. The horizontal component of Fig. 6.48 weight (i.e., force) along the direction of motion of the bob is given by Fx = – mg cos  Negative sign shows that Fx is directed towards mean position O. This force is restoring force. x cos  = R , where R is the radius of Earth. x  Fx = – mg R Now, acceleration produced in the bob of pendulum, a = Fx = – g x m  R  g Since  R  is constant.  a  – x So, motion of the pendulum is SHM. Time period of the pendulum T = 2 = 2 = 2 Now R = 6.4 × 106 m; g = 9.8 m s–2  T = 2 × 3.14  s = 5075 s = 84.58 min. ~ 84.6 minutes.

314 Physics—XI 29. Two small masses m1 and m2 are attached to the ends of a rod of negligible mass as shown in Fig. 6.49. The rod is pivoted so that the masses and rod form a physical pendulum. Find an expression for the period of oscillation. Ans. When the system is rotated about the pivot through an angle , then restoring torque,  = m2g l2 sin  – m1gl1 sin  or  = (m2g l2 – m1g l1)  [... sin   , for small ] or I = (m1l12 + m2l22) = (m2gl2 – m1gl1) T = 2  = 2 m1l12  m2l22 .  m2gl2  m1gl1 Fig. 6.49 30. Two simple harmonic motions are represented by the following equations : y1 = 10 sin  (12t + 1) 4 y2 = 5 (sin 3 t + 3 cos 3 t) Find out the ratio of their amplitudes. What are the time periods of two motions?     = 10 sin      ...(1) Ans. y1 = 10 sin     Comparing eq. (1) with standard equation of SHM, y = A sin (t + 0) then, A1 = 10, 1 = 3  T1 =  = y2 = 5 sin 3t + 5 cos 3t ...(2)  ...(3) Let 5 = A2 cos  ...(4) ...(5) and 5 = A2 sin   y2 = A2 cos  sin 3t + A2 sin  cos 3t Squaring and adding = ns (3) and (4), we get A22 = 52 + (5 )2 or A2 = 10 ; T2 =  =  A1 = 10 = 1. A 0.06 m 0.06 m B A2 10 P /6 /6 Q 31. Two identical balls A and B, each of mass 0.1 kg, are attached to two identical massless springs. The spring- Fig. 6.50 mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in Fig. 6.50. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 m. Each spring has a natural length of 0.06  metre and spring constant 0.1 N m–1. Initially, both the balls are displaced by an

Oscillations 315 angle  =  radian w.r.t. the diameter PQ of the circle and released from rest. 6 (i) Calculate the frequency of oscillations of ball B. (ii) What is the total energy of the system? (iii) Find the speed of ball A, when A and B are at the two ends of the diameter PQ. Ans. Effective force constant, k = (0.1 + 0.1) N m–1 = 0.2 N m–1 Reduced mass,     (i) Frequency,         (ii) Maximum compression or elongation,        or   = 0.02  m Total energy of system, E      = 3.95 × 10–4 J (iii) If v is the required velocity, then   = 2  3.95  10–4 m s–1 = 12.57 × 10–2 m s–1.  0.05 32. If a tunnel be dug inside the Earth (not passing through the centre of Earth) and a ball be dropped at one end of it, then show that the ball will execute simple harmonic motion. Also find its time period. Given : Radius of Earth = 6.4 × 106 m and g = 9.8 m s–2.   F Ans. F = – Fg cos  = – mg    OP F = – mg = –   Fg F=– y C F=–     y   F = – Gmy Fig. 6.51 Clearly, k = Gm Now, T = 2 =     But  =  T= = 2  = 2 = 2 × 3.14  s = 5076 s = 84.6 minutes. = 2

316 Physics—XI 33. A child suspends a circular hoop of mass M and of radius R R from a nail so that it hangs in a vertical plane as shown in  Fig. 6.52. Find the period of this physical pendulum for small oscillations about equilibrium. Fig. 6.52 Ans. The moment of inertia for a hoop about its axis is ICM = MR2. The pivot point, however, is not along the axis of the hoop, so we must apply the parallel axis theorem. I = ICM + Md2 In the given problem, the distance between the axis through the centre of mass and the pivot point is just the radius R of the hoop.  I = MR2 + MR2 = 2MR2 The distance between the pivot and the centre of mass is R.  T = 2 = 2 = 2 This period is equal to that of a simple pendulum with length 2R. O  l 34. A ball is suspended by a thread of length l at the point O on the wall, forming a small angle  with the vertical. Then the thread with the ball was deviated through a small angle  ( > ) and set free. Assuming the collision of the ball against the wall to be perfectly elastic, find the oscillation period of such a pendulum. Ans. Total time taken from OB to OC and back to OB    where  Fig. 6.53 Again, let t be the time taken from OB to OA. The time taken to O go from OA to OB is also t.  l Now,     ;    or               Total time             A C    B          Fig. 6.54 35. Dpoeuterremd iinnetothaebpeenrtiotudboef[oFsicgi.ll6a.t5i5o]nws hofosmeerricguhrtyaorfmmfaosrsmms a=n2a0n0ggle   = 30° with the vertical. The cross-sectional area of the tube is Fig. 6.55 S = 0.50 cm2. The viscosity of mercury is to be neglected. Ans. 0.8 s Hint. Let the level of mercury be depressed in the vertical arm of the tube through a distance y. Then, vertical rise of mercury in the slant portion of the tube = y cos  Vertical difference in liquid level in two arms =     Restoring force F = – [Weight of mercury column of length y (1 + cos)]

Oscillations y ycos  317 or      where  is the density and S is the cross-sectional area of the tube. Comparing with F = – ky,     Period of oscillation,  or    . Fig. 6.56 36. A body of mass m fell from a height h on to the pan of a spring balance. The masses of the pan and the spring are negligible, the stiffness of the latter is k. Having stuck to the pan, the body starts performing harmonic oscillations in the vertical direction. Find the amplitude and the energy of these oscillations. m Ans. Suppose the spring is compressed through a distance x. h Loss of gravitational potential energy = mg (h + x) Gain of elastic potential energy of the spring  k Equating,  Fig. 6.57 or    or              is the distance through which the spring is compressed in the equilibrium position. Amplitude of oscillation, A =  Energy of oscillation                   37. The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (t + ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is  cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is  s–1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (t + ), what are the amplitude and initial phase of the particle with the above initial conditions. Ans. At t = 0, x = 1 cm and v =  cm s–1, Using x = A cos (t + 0), 1 = A cos 0 ...(1)

318 Physics—XI Now, v = = – A sin (t + 0)  = – A  sin 0 or A sin 0 = – 1 ...(2) Squaring and adding (1) and (2), A2 (cos2 0 + sin2 0) = 1 + 1 = 2 ...(3) or A2 = 2 or A = cm ...(4) Dividing (2) by (1), tan 0 = – 1 or 0 =   or Again, x = B sin (t + ) At t = 0, x = 1 and v =   1 = B sin  v = B cos (t + ) and  = B cos  or 1 = B cos  From (3) and (4), B2 (cos2  + sin2 ) = 1 + 1 = 2 or B= Dividing (3) by (4), we get tan  = 1 or  =  or  . 38. Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case : (x is in cm and t is in s). (a) x = –2 sin (3t +  / 3 ) (b) x = cos (/6 – t) (c) x = 3 sin (2 t +  / 4 ) (d) x = 2 cos  t Ans. Let us express each function in form x = A cos (t + 0). Then A represents radius of the reference circle,  is angular speed of the rotating particle and  is the angle the initial radius vector of the particle makes with the positive direction of x-axis. (a) x =     or x =                Comparing with x = A cos (t + 0), we get A = 2,  = 3 and      =  The reference circle is as shown in Fig. 6.58. (b) x =     or x=        [... cos (– ) = – cos ] Comparing with x = A cos (t + 0), A = 1,  = 1,     Fig. 6.58 The reference circle is as shown in Fig. 6.59. (c) x =   ; x =                  Comparing with x = A cos (t + 0), A = 3,  = 2,      or 0 =  The reference circle is as shown in Fig. 6.60. Fig. 6.59

Oscillations 319 (d) x = 2 cos t Comparing with x = A cos (t + 0), A = 2,  =  and 0 = 0 The reference circle is as shown in Fig. 6.61. Fig. 6.60 Fig. 6.61 39. Figure 6.62 shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 6.63 shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 6.63 is stretched by the same force F. km m k m F F F Fig. 6.62 Fig. 6.63 (a) What is the maximum extension of the spring in the two cases? (b) If the mass in Fig. 6.62 and the two masses in Fig. 6.63 are released free, what is the period of oscillation in each case? Ans. (a) Maximum extension  (for both cases) (b) In Fig. 6.62, m is the inertia factor and k is the spring factor. T = 2 or T = 2 In Fig. 6.63,   Subtracting,  = – kx – kx ...(1) Change in length of spring,      [Note that l is constant] From equation (1),   or  or   or  

320 Physics—XI or      Aliter. Reduced mass of the given two-body system, =    inertia factor = and spring factor = k   Time period, T = 2 = 2 . 40. Answer the following questions : (a) Time period of a particle in SHM depends on the force constant k and mass m of the particle :  A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum? (b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π l g . Think of a qualitative argument to appreciate this result. (c) A man with a wrist watch on his hand falls from the top of a tower. Does the watch give correct time during the free fall? (d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity? Ans. (a) For a simple pendulum, k itself is proportional to m. So, m cancels out. (b) Sin  <  ; if the restoring force, mg sin  is replaced by mg , this amounts to effective reduction in g for large angles, and hence an increase in time period T over that given by the formula T =  , where one assumes sin  =  (approximately true for same angles). (c) Yes, because the wrist watch depends on spring action and has nothing to do with acceleration due to gravity. (d) Gravity disappears for a man under free fall. So, frequency is zero. 41. An air chamber of volume V has a neck area of cross-section a into which a ball of mass m just fits and can move up and down without any friction as shown in figure. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be m isothermal [see figure]. a Ans. Consider a frictionless ball of mass ‘m’ fitted smoothly in the V neck of an air chamber of volume V. Let the ball be pushed down through a small distance y and released. The ball will begin to Air execute simple harmonic motion. The cross-sectional area of the neck is a. If V be the decrease in Fig.6.64 volume when the ball is pressed down, then V = a y

Oscillations 321 Volumetric strain =  y Let p be the excess pressure, i.e., hydrostatic stress V produced due to the pressing down of the ball. Fig. 6.65 Bulk modulus of elasticity, law of motion]  B = or B =  Negative sign indicates that the volume decreases as the stress increases.  Now, p =   Force, F =    a or F =    or F =  or m =  [Newton’s 2nd or  or  =0 Putting = 2, we get  =0 which is the equation of simple harmonic motion. Thus, the ball executes simple harmonic oscillation. Now, 2 = or   or 2 = or  =  Time period, T =   .  42. You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg. Take g = 10 m s–2. Ans. (a) M = 3000 kg, x = 0.15 m Spring constant of the parallel combination of four springs = 4 k, where k is the spring constant of each spring. Now, Mg = 4kx or k =  N m–1 = 5 × 104 N m–1 

322 Physics—XI (b) Mass supported by each spring, m = kg = 750 kg or = 2 Using  = x= , or loge = loge 2 or loge e = loge 2 or = loge 2 or b = Now, T = 2 =2×   s = 0.7698 s   kg s–1 = 1350.5 kg s–1.  b= 43. Show that for a particle in linear SHM, the average kinetic energy over a period of oscillation equals the average potential energy over the same period. Ans. y =  , v=   Kinetic energy, Ek =   Potential energy, Ep =   Average kinetic energy over one cycle =  =          =   =                 =           =       ...(1)   Average potential energy over one cycle =       =        

Oscillations 323 =        ...(2)  It follows from (1) and (2) that the average kinetic energy over a period of oscillation equals the average potential energy over the same period. 44. A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm. Ans. A = 5 cm, T = 0.2 s,  =  =  = 10 rad s–1 Now, a = – 2y and v =   (a) a = –   y=–  × m s–2 = – 5 2 m s–2    v=    m s–1 = 0 = (b) a = –  y = –  × 3 × 10–2 m s–2 = – 3 2 m s–2   × 10–2 m s–1 = 0.4  m s–1   v= = (c) a = –  y = –  × 0 = 0 v =    × 10–2 m s–1 = 0.5  m s–1. = 45. A mass attached to a spring is free to oscillate, with angular velocity  , in a horizontal plane without friction or damping. It is pulled to a distance x 0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters  , x0 and v0 . Ans. x = A cos (t + ), v = = – A sin (t + ) When t = 0, x = x0 and = – v0  x0 = A cos  ...(i) and – v0 = – A sin  or A sin  = ...(ii)  Squaring and adding (i) and (ii), we get A2 (cos2  + sin2 ) = x02 +  or A = .  46. A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s–1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time. (a) Will there be any change in weight of the body, during the oscillation? (b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position? Ans. (a) Yes. When the platform is accelerating upwards with acceleration a, then the effective weight of the person is m(g + a). When the platform is accelerating downward with acceleration a, then the effective weight of the person is m(g – a).

324 Physics—XI (b) Maximum acceleration of platform, a = 2A = 422A = 4(3.14)2 × 22 × 5 × 10–2 m s–2 = 7.89 m s–2 Maximum reading in machine while moving upwards = m(g + a) = 50(9.8 + 7.89) = 884.5 N Minimum reading in machine while moving downwards = 50(9.8 – 7.89) = 95.5 N Maximum weight is at the topmost position. Minimum weight is at the lowermost position. 47. A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion. Ans. Let g be the value of acceleration due to gravity at P. P is at a distance d below the surface of Earth. R is the radius of Earth.    A   g =   y=R–d g= g × y d P y=R–d O R Force on body at P, F = mg = This force is directed towards mean position O. Fig. 6.66 Moreover, F y. So, body executes SHM. Spring factor, k = Inertia factor = m and T =  T=   CASE STUDY BASED QUESTIONS Read the following passages and answer the questions that follow. CASE STUDY 1 Pendulum clocks are often seen at railway stations, old churches and Museums. What makes these clock work? These clocks use a torsion pendulum as a swinging weight (circular bob), which serves as a crucial time-keeping element. The pendulum is hung from its centre of the clock with the help of a wire torsion spring due to which it is rferred as torsion pendulum. The weight oscillates along the axis of the clock under the effect of the torsion spring. The spring is twisted and reversed, which causes the oscillation in both clockwise and counterclockwise direction. 1. A pendulum is released from the horizontal position Fig. 6.67 as shown in the given figure. The motion is

Oscillations 325 (a) S.H.M. (b) Periodic but not S.H.M. (c) aperiodic but S.H.M. (d) Oscillatory but aperiodic Ans. (b) 2. What is the length of a pendulum for which time period is 1s at a place where g = 9.8 ms–2? (a) 14.8 cm (b) 22.8 cm (c) 24.8 cm (d) 32.2 cm Ans. (c) Hint and Solution: T = 2 L , L  gT 2  9.8  (1)2 g 42 42 3. The time period of a simple pendulum varies with the mass of the bob (m) as (a) m0 (b) m1/2 (c) m (d) m–1/2 Ans. (a) Hint and Solution: T = 2 L. g Time period of a simple pendulum does not depends upon mass of the bob. 4. The bob of a simple pendulum of length l is negatively charged. A positively-charged metal plate is placed just below the bob and the pendulum is made to oscillate. What will be the effect on the time-period of the pendulum? (a) increases (b) decreases (c) remains same (d) either (a) or (c) Ans. (b) Hint and Solution: The positively charged metal plate attracts the negatively charged bob. It increases the effective value of g, hence the time period will decrease. CASE STUDY 2 Samit performs an experiment, he takes a liquid in a U-tube and maintain equal level in both the columns of the tube. When air is blown in on side B, the liquid may go down upto point D. The displacement of the liquid is given by BD = x. In the second column, the liquid goes up to E and the displacement is CE = x. As the applied force is withdrawn, the liquid level in the second column comes down because of gravity from E to C. But due to inertia, it overshoots the mark C and goes to F (CF = x), whereas the liquid in the other column also goes up by the Fig. 6.68 same distance. Thus, the liquid oscillating in both the columns. 1. A mercury column of mass m oscillates in a U-tube. One centimetre of the mercury column weighs 12 g. The spring constant (k) of motion is (a) 23.52 Nm–1 (b) 23.25 Nm–1 (c) 24.8 Nm–1 (d) 26.32 Nm–1 Ans. (a)

326 Physics—XI Hint and Solution: If the mercury in right arm be depressed through a distance x metre. Then the liquid in the left arm will rise through a distance x metre. Restoring force (F) = – Weight of unbalanced column of length 2 x metre F= 12  103  9.8  2x = –23.52 y 1  102 Comparing with F = ky, we get k = 23.52 Nm–1 2. Motion of an oscillating liquid column in a U-tube is (a) periodic but not simple harmonic (b) non-periodic (c) Simple harmonic and time period is independent of the density of the liquid (d) Simple harmonic and time period is directly proportional to the density of the liquid Ans. (c) Hint and Solution: T = 2 L, where l is the height of liquid column in one arm g of U-tube. 3. The physical quantity which remains constant in simple harmonic motion is (a) kinetic energy (b) displacement (c) restoring force (d) frequency Ans. (d) 4. A vertical U-tube of uniform cross-section contains water upto a height of 2.4 cm. When the water on one side is depressed and then released, its up and down motion in the tube is observed. The time period of the motion is (Given g = 980 cms –2)  2 3 (d) none of these (a) (b) (c) 10 10 10 Ans. (a) Hint and Solution: T = 2 L  2 2.45 s   g 980 10 CASE STUDY 3 When a particle of mass m moves on the x-axis in a potential of the form V( x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to , as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = x4 ( > 0) for |x | near the origin and becomes a constant equal to V0 for | x |  X0 (see figure).

Oscillations 327 Fig. 6.69 1. If the total energy of the particle is E, it will perform periodic motion only if (a) E < 0 (b) E > 0 (c) V0 > E > 0 (d) E > V0. Ans. (c) Hint and Solution: Energy must be less than V0. 2. For periodic motion of small amplitude A, the time period T of this particle is proportional to (a) A (b) (c) A  (d) .   Ans. (b) Hint and Solution: Only (b) has the dimensions of time. 3. The acceleration of this particle for | x | > X0 is (a) proportional to V0 (b) proportional to (c) proportional to (d) zero. Ans. (d) Hint and Solution: As potential energy is constant for |x| > x0, the force on the particle is zero. Hence acceleration is zero. 4. The equation of a damped simple harmonic motion is m + b + kx = 0. Then the angular frequency of oscillation is   (b) =    (a)  =       (c)  =    (d)  =  .  Ans. (a)

Chapter 7: Waves (NCERT Textbook Chapter-15) SUMMARY OF THE CHAPTER • A wave motion is a kind of disturbance that travels through a medium due to the repeated periodic motion of the particles of the medium about their mean positions, the motion gets handed over from one particle to another. • The phenomenon of wave propagation on the water surface can be explained on the basis of elasticity and inertia of the medium. • The elastic or mechanical waves are those waves which can be propagated only through a meterial medium. On the other hand, non-mechanical or electromagnetic waves do not require any material medium for their propagation. For example, sound waves and waves on a water surface are mechanical waves but light waves and X-rays are non-mechanical waves. • The wave motion is of two types: longitudinal and transverse. In case of a transverse wave motion, the particles of the medium vibrate about their mean positions in a direction perpendicular to the direction of propagation of disturbance. However, in case of a longitudinal wave motion, the particles of the medium vibrate about their mean positions in the direction of propagation of disturbance. • A transverse wave travels in the form of crests and troughs. The crest refers to that portion of the medium, which is highly raised above the normal positions of the particles of the medium as the wave passes through it. The trough refers to that portion of the medium, which is highly depressed below the normal positions of rest of the particles of the medium, as the wave passes through it. • A longitudinal wave travels in the medium in the form of compressions and rarefactions. The compression refers to that region of the medium, in which particles of the medium come to distances less than the normal distance between them. On the other hand, the rarefaction refers to that region of the medium, in which particles of the medium move apart to distances greater than the normal distance between them. • The distance between two successive crests or successive troughs is called wavelength. The distance between two successive compressions or successive rarefactions is also called wavelength. • The velocity (v) of longitudinal waves is given by, v =  where  is frequency and  is wavelength. • The speed of transverse waves in a solid is given by, v=  where E is elasticity of the medium and  is its density. 328

Waves 329 • The speed of transverse waves in a stretched string is given by, v= where T is tension in the string and m is mass per unit length, also called linear density of the string. • According to Newton, the speed of sound in a gaseous medium is given by, v=  where P is pressure of the gas and  is its density. • According to Laplace’s correction, the speed of sound in a gaseous medium is given by,  v =  , where  = • The speed of sound in a gas is inversely proportional to the square root of the density of the gas. • The speed of sound varies directly as the square root of temperature but is independent of the changes in pressure. The velocity of sound increases nearly by 0.61 m s–1 for each 1 °C rise of the temperature. • The displacement of the wave pulse at a point at distance x from the origin at t = 0, is given by y = f(x) For a wave pulse travelling with velocity v from left to right, the wave function is y(x, t) = f(x – vt) But if the wave is travelling from right to left, then the wave function is y(x, t) = f(x + vt) • The waves represented by the wave function     y(x, t) = a sin   are known as harmonic wave. Here ‘a’ is called amplitude. • The wave travelling continuously in a medium in the same direction without any change in its amplitude is known as a progressive wave or a travelling wave. The progressive wave may be transverse or longitudinal in nature. • According to the principle of superposition of waves, when two or more waves propagate in a medium in such a way that each wave represents its individual motion separately, then the resultant displacement of any particle of the medium at any instant is equal to the vector sum of the individual displacements i.e.        • When two waves of same frequency (or wavelength) moving with same speed in the same direction in a medium superpose each other, they give rise to interference.

330 Physics—XI • When two waves of same frequency moving with same speed in opposite directions in a medium superpose each other, they give rise to stationary waves. • When two waves of slightly different frequencies moving with the same speed along the same direction superimpose each other, they give rise to beats. • The wavelength of nth mode of vibration of the stretched string (length L) is given by, n = and the frequency is given by, n = n = n 1, the symbols have their usual meanings. It is the (n – 1)th overtone or nth harmonic. • In case of a pipe open at both ends (L being the length of the pipe), the frequency of the nth mode of vibration is given by, n = n = n 1, the symbols have their usual meanings. It is (n – 1)th overtone or nth harmonic. Important Results 1. Wave velocity = Frequency × wavelength or v =  2. Wave velocity =  or v = 3. Newton’s formula for the speed of sound in a gas, v =   4. Laplace’s formula for the speed of sound in a gas, v =  For air  =  5. Velocity of sound in a gas, v =  . When pressure changes, density also changes in the same ratio, hence  remains unchanged. So, change in pressure has no effect on the speed of sound in a gas. 6. Speed of sound is inversely proportional to the square root of density. v  7. Speed of sound in a gas is directly proportional to the square root of its absolute temperature. v

Waves 331 8. The fundamental frequency of vibration of a stretched string is given by,  =  , =  ×    =   or  =  where L is first resonance length, D is diameter of the string, T is tension in the string and  is density of the material of the string. 9. Fundamental frequency, 1 = . It is the lowest frequency of vibration producing fundamental note or first harmonic and L is first resonance length. 10. n = n 1 The frequency of nth harmonic is n times the fundamental frequency. The nth harmonic is also called (n – 1)th overtone. 11. In case of closed organ pipes, Fundamental frequency, 1 = where v is velocity of sound in air and L is length of the pipe. First overtone or 3rd harmonic, 2 = 3 1 Second overtone or 5th harmonic, 3 = 51. In general, the frequency of note produced in nth normal mode of vibration of closed organ pipe is given by, n = (2n – 1) 1 It is (2n – 1)th harmonic or (n – 1)th overtone. It must be noted that only alternate harmonics of frequencies 1, 31, 51,... etc., are present. Harmonics are the notes/ sounds of frequency equal to or an integral multiple of fundamental frequency. 12. In case of open organ pipe, the frequency of vibration in nth normal mode of vibration is given by, n = n1 The note so produced is nth harmonic or (n – 1)th overtone. It must be noted that in an open organ pipe, all the harmonics are present. 13. In case of resonance apparatus, v = 2(l2 – l1)  end correction, e = Also, e = 0.3 D (theoretically), where D is diameter of the resonance tube. 14. The number of beats produced per second or beat frequency is equal to the difference in the frequencies of the interfering waves. 15. A plane progressive harmonic wave travelling along positive direction of X-axis can be represented by any of the following expressions: (i) y = a sin (t – kx), k = 2/    (ii) y = a sin 2     (iii) y = a sin (vt – x) 

332 Physics—XI where  is the wavelength, v is the velocity, A the amplitude and x is the distance of observation point from the origin. 16. For a progressive wave travelling along –ve X-axis, y = a sin (t + kx) or    = a sin  y = a sin 2    (vt + x)  17. Phase,    + 0, where0 is the initial phase.  = 2    18. Displacement y and amplitude a have same units of length. 19. Beat frequency (b) = Number of beats per second = Difference in frequencies of two sources or b = (1 – 2) or (2 – 1) 20. 2 = 1  b 21. If the prong of tuning fork is filed, its frequency increases. If the prong of a tuning fork is loaded with a little was, its frequency decreases. VERY SHORT ANSWER TYPE QUESTIONS (1 Mark) 1. A boy stands 66.4 m in front of a building and then blows a whistle (Fig. 7.1). Calculate the time interval when he hears an echo. (Take speed of sound = 332 m s–1). Ans. Time = = Fig. 7.1 s = 0.4 s. 2. What is the range of frequency of audible sound? Ans. 20 Hz to 20 kHz. 3. Why does sound travel faster in iron than in air? Ans. It is because solids are highly elastic as compared to gases. 4. What kind of waves help the bats to find their way in the dark? Ans. Ultrasonic waves. 5. The velocity of sound in air is 332 m s–1. Find the frequency of the fundamental note of an open pipe 50 cm long. Ans.  = = Hz = 332 Hz.

Waves 333 6. In which gas, hydrogen or oxygen, will sound have greater velocity? Ans. As v  , therefore velocity of sound will be greater in hydrogen gas.  7. The figure 7.2 shows two situations in which the same string is put under tension by a suspended mass of 5 kg. In which situation will the speed of waves sent along the string be greater? Fig. 7.2 Ans. (a). This is due to larger tension  . 8. In a resonance tube, the second resonance does not occur exactly at three times the length at first resonance. Why? Ans. This is due to end-correction. 9. The frequency of the fundamental note of a tube closed at one end is 200 Hz. What will be the frequency of the fundamental note of a similar tube of the same length but open at both ends? Ans. 400 Hz. 10. A wave transmits energy. Can it transmit momentum? Ans. Yes. 11. A string has a linear density of 0.25 kg m–1 and is stretched with a tension of 25 N. What is the velocity of the wave? Ans.    . 12. By how much the wave velocity increases for 1°C rise of temperature? Ans. 61 cm s–1. 13. In a longitudinal wave, what is the distance between a compression and its nearest rarefaction?  Ans. 14. The ratio of the amplitudes of two waves is 3 : 4. What is the ratio of the intensities of two waves? Ans.    15. When a vibrating tuning fork is moved speedily towards a wall, beats are heard. Why? Ans. This is due to difference in the frequency of the incident wave and the apparent frequency of the reflected wave. 16. How does the speed of sound in air vary with temperature?

334 Physics—XI Ans. It increases directly as the square root of the absolute temperature of air i.e., v  . 17. Two sound sources produce 12 beats in 4 s. By how much do their frequencies differ? Ans. Number of beats per second = = 3  1 – 2 = 3. 18. Why sound is heard with more intensity through carbon dioxide than through air? Ans. This is because the intensity of sound increases with the increase in the density of the medium. 19. Classify the following waves into longitudinal and transverse. Waves on the surface of water, Sound waves in air, Electromagnetic waves, wave set up on the wire of a sitar. Ans. In the given list, only the sound waves are longitudinal. 20. Why transverse waves cannot be set up in a gas? Ans. This is because a gas has no rigidity. 21. Is it possible to have interference between the waves produced by two violins? Ans. No. 22. Two waves of the same frequency and same amplitude, on superposition, produce a resultant wave of the same amplitude. What is the phase difference between the two waves?  Ans. 23. What is the relation between path difference and phase difference? Ans. Phase difference =   Path difference.  24. What is the shape of the graph between the pressure of a gas and the velocity of sound waves passing through the gas? Ans. It is a straight line parallel to the pressure axis. 25. The velocity of sound in hydrogen is.... times as compared to the velocity of sound in oxygen at constant temperature.   Ans. four.     .  26. The window panes of houses sometimes get cracked due to some explosion at large distance. Which waves are responsible for this? Ans. Shock waves. 27. What is the effect of pressure on the velocity of sound waves? Ans. There is no effect. 28. Define persistence of hearing. Ans. It is the duration of time for which a syllable’s impression remains in the ear. 29. If the absolute temperature of a gas is quadrupled, then what is the effect on the velocity of sound in that gas? Ans. Since  therefore the velocity of sound will be doubled.

Waves 335 30. At the same temperature and pressure, the densities of two diatomic gases are d1 and d2 . What is the ratio of the speeds of sound in these gases? Ans.  . 31. What will be the effect on the frequency of sonometer wire if the tension is decreased by 2%? Ans. The frequency shall decrease by 1%. 32. What is the distinguishing feature between a sound and its echo? Ans. Loudness. 33. The velocity of sound in a tube containing air at 27°C and a pressure of 76 cm of mercury is 330 m s–1. What will be the velocity of sound when pressure is increased to 100 cm of mercury and the temperature is kept constant? Ans. 330 m s–1. This is because the velocity of sound is independent of pressure. 34. Why bells are made of metal and not wood? Ans. This is because wood has high damping. 35. What is the distinguishing factor between two notes emitted by two different tuning forks? Ans. Pitch. 36. What is the distinguishing factor between two notes emitted by two wires of different materials having the same fundamental frequency? Ans. Quality. 37. Two strings of equal length are stretched by equal forces. The strings are made of the same materials. If their diameters are in the ratio 3 : 2, what will be the ratio of their fundamental frequencies? Ans.     .  38. The diameter of a stretched wire is halved. What would be the effect on the wave speed? Ans. It would be doubled. 39. The frequency of the first overtone of a closed organ pipe is the same as that of the first overtone of an open pipe. What is the ratio between their lengths? Ans. 3 : 4. 40. An observer places his ear at the end of a long steel pipe. He can hear two sounds, when a workman hammers the other end of the pipe. Why? Ans. This is because sound is transmitted both through air and medium. 41. Why the velocity of sound is generally greater in solids than in gases? Ans. This is because for solids is much greater than for gases.  42. Why no beats can be heard if the frequencies of the two interfering waves differ by more than ten? Ans. This is due to persistence of hearing. 43. Why a stationary wave is so named? Ans. A stationary wave is so named because there is no net propagation of energy.

336 Physics—XI 44. Two identical sound waves pass through a medium at a point with a phase difference of 180°. Whether the interference at that point will be constructive or destructive? Ans. Destructive. 45. Why is it not possible to have interference between the waves produced by two violins? Ans. This is because the sounds produced will not have a constant phase relationship. 46. How do we identify our friend from his voice while sitting in a dark room? Ans. The quality of sound helps us to identify the sound. 47. The distance between two consecutive nodes in a stationary wave is 15 cm. If the speed of the wave be 300 m s– 1, calculate the frequency. Ans.     ,     48. The frequencies of two tuning forks A and B are 250 Hz and 255 Hz respectively. Both are sounded together. How many beats will be heard in 5 second? Ans. 1 = 250 Hz ; 2 = 255 Hz Number of beats per second, m = 255 – 250 = 5 Number of beats in 5 second = 25. 49. An observer standing at a sea-coast observes 54 waves reaching the coast per minute. If the wavelength of the waves is 10 m, find the velocity of the waves. Ans.     ,       9 m s–1 50. Deduce the velocity of longitudinal waves in a metal rod. Given : modulus of elasticity = 7.5 × 1010 N m–2 and density = 2.7 × 103 kg m–3. Ans. v =  =  m s–1 = 5.27 × 103 m s–1  SHORT ANSWER TYPE–I QUESTIONS (2 Marks) 1. A tuning fork of unknown frequency gives 4 beats with a tuning fork of frequency 310 Hz. It gives the same number of beats on filing. Find the unknown frequency. Ans. Out of the two possible frequencies 314 Hz and 306 Hz, one is the initial value and the other is the final value. Since the frequency increases on filing therefore the initial frequency is 306 Hz. 2. The string of a violin emits a note of 540 Hz at its correct tension. The string is bit taut and produces 4 beats per second with a tuning fork of frequency 540 Hz. Find the frequency of the note emitted by this taut string. Ans. The frequency of vibration of a string increases with increase in the tension. Thus, the note emitted by the string will be a little more than 540 Hz. As it produces 4 beats per second with the 540 Hz tuning fork, the frequency emitted by the taut string will be 544 Hz.

Waves 337 3. Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both : (i) Motion of a kink in a long coil spring produced by displacing one end of the spring sideways. (ii) Waves produced in a cylinder containing a liquid by moving its piston back and forth. (iii) Waves produced by a motorboat sailing in water. (iv) Ultrasonic waves in air produced by a vibrating quartz crystal. [NCERT Solved Example] Ans. (i) Transverse (ii) Longitudinal (iii) Transverse and longitudinal (iv) Longitudinal. 4. The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of sound in air is 330 ms–1. Find the length of the air column. [End correction may be neglected] Ans. L = =  m = 0.94 m.   5. In the following series of resonant frequencies, one frequency (lower than 400 Hz) is missing : 150, 225, 300, 375 Hz (a) What is the missing frequency? (b) What is the frequency of the seventh harmonic? Ans. (a) 75 Hz. [If 75 Hz is taken as , then we have 2, 3, 4, 5, ...] (b) Frequency of 7th harmonic = 7 × 75 Hz = 525 Hz. 6. Flash and thunder are produced simultaneously. But thunder is heard a few second after the flash is seen. Why? Ans. This is because the speed of light is much larger than the speed of sound. 7. Plot a graph between pressure and velocity of sound in a gas. Ans. The velocity of sound in a gas is independent of pressure, provided temperature remains constant. v P Fig. 7.3 8. Name two properties which are common to all types of mechanical waves. Ans. Mechanical waves travel in a medium. The medium itself does not move with the wave. 9. The fundamental frequency of an organ pipe is 130 Hz. The frequency of the first overtone is 390 Hz. Identify the organ pipe.

338 Physics—XI Ans. Closed organ pipe. This is because the frequency of the first overtone is three times the fundamental frequency. 10. How does a balloon filled with CO2 gas behave for sound as a lens? The balloon is given to be surrounded by air. Ans. The balloon will behave as a convex lens. This is because the velocity of sound in CO2 is less as compared to that in air. So, CO2 acts as a denser medium as compared to air. 11. Which physical quantity is represented by the ratio of intensity of wave and energy density? Ans. Intensity =  , Energy density =    = velocity.   12. What is the longest wavelength possible when the air column in a closed pipe of length 20 cm vibrates? Ans. For longest wave-length, we need to consider the fundamental mode of vibration.  or  = 80 cm = 0.80 m. AN /4 Fig. 7.4 13. What is the longest wavelength that can produce a stationary wave in an open tube of length 10 cm? Ans. For longest wavelength, we shall consider the fundamental note  or  = 20 cm = 0.20 m. A A N /2 Fig. 7.5 14. When are the tones called harmonics? Ans. The tones are called harmonics if the frequencies of the fundamental tone and other overtones produced by a source of sound are in harmonic series.

Waves 339 15. In sound, beats are heard when two independent sources are sounded together. Is it possible in the case of sources of light? Ans. In the case of light, it is not possible. This is because the phase relationship between two independent light sources is random. 16. Name two factors which determine the pitch of a tuning fork. Ans. The pitch of tuning fork is determined by frequency. The frequency is inversely proportional to the square of the length and directly proportional to the thickness of the fork. 17. What will be the velocity of sound in a perfectly rigid rod and why? Ans. The velocity of sound in a perfectly rigid rod will be infinite. This is because the value of Young’s modulus of elasticity is infinite for a perfectly rigid rod. 18. Sound is simultaneously produced at ‘one ends’ of two strings of the same length, one of rubber and the other of steel. In which string will the sound reach the other end earlier and why? Ans. The value of is larger for steel as compared to rubber. So, sound would reach the other end earlier in the case of steel string. 19. What will be the effect on the frequency of the sonometer wire if the load stretching the sonometer wire is immersed in water? Ans. Due to the upthrust experienced by the load, the effective weight shall decrease. The tension shall decrease. So, the frequency shall decrease. 20. The weight suspended from a sonometer wire is increased by a factor of 4. Will the frequency of the wire be increased exactly by a factor of 2? Justify your answer. Ans. No. There will a slight increase in the length of the wire. So, the frequency shall become slightly less than double. 21. The frequencies of the two sources of sound are 512 Hz and 516 Hz. What is the time interval between two consecutive maxima produced by sounding them together? Ans. Required time interval is the reciprocal of the difference in frequencies      .   22. An explosion takes place at the bottom of a lake. Will the shock waves in water be longitudinal or transverse? Ans. Shock wave is a longitudinal wave. However, it travels with a speed which is greater than the speed of a longitudinal wave of normal intensity. 23. Why the bells of colleges and temples are of large size? Ans. Larger the area of the source of sound, more is the energy transmitted into the medium. Consequently, the intensity of sound is more and loud sound is heard. 24. An organ pipe is in resonance with a tuning fork. If air of the organ pipe is replaced by hydrogen, then how should the length of the pipe be changed for resonance? Ans. The velocity of sound shall increase. The frequency shall increase. In order to lower the frequency, the length of the pipe shall have to be increased. 25. An organ pipe is in resonance with a tuning fork. If the temperature is increased, then how should the length of the pipe be changed for resonance?

340 Physics—XI Ans. When the temperature increases, the velocity of sound increases. The frequency increases. In order to reduce the frequency, length shall have to be increased. 26. A travelling wave in a stretched string is described by the equation y = A sin (kx – t). What is the maximum particle velocity? Ans. Particle velocity = (y) = [A sin (kx – t)] = – A cos (kx – t)  Maximum particle velocity = A. 27. Show that the fundamental frequency of an open organ pipe is twice the fundamental frequency of the closed organ pipe of the same length. Ans. Fundamental frequency of an open organ pipe, = Fundamental frequency of a closed organ pipe,  =  or  = 2.  = × = 2 28. An organ pipe is in resonance with a tuning fork. If pressure of air in the pipe is increased by a factor of 1.3, then how should the length be changed for resonance? Ans. The velocity of sound is independent of pressure. So, there is no change in frequency. Thus, there is no need to change the length of the pipe. 29. The distance between two consecutive nodes in a stationary wave is 25 cm. What is the frequency of the wave if the velocity of wave is 300 m s–1?   , Now,    . Ans.   30. A resonance tube resonates with a tuning fork of frequency 256 Hz. If the lengths of the resonated air column are 32 cm and 100 cm, what is the value of the end correction? Ans. l1 + e =     ,Now, 3l1 + 3e =  Now, 3l1 + 3e = l2 + e or     or e =  . 31. In the previous question, what is the speed of sound? Ans. Speed of sound =  =  × 2 (l2 – l1) =    . 32. What is the ratio of speeds of sound in hydrogen and oxygen at the same temperature? Ans.  ,  . 

Waves 341 33. If two persons are talking on the surface of Moon, they cannot hear each other. Why? Ans. Sound waves require material medium for propagation. Since there is practically no atmosphere on the Moon therefore the two persons cannot hear each other. 34. If oil of density higher than that of water is used in place of water in a resonance tube, how does the frequency change? Ans. The frequency is governed by the air column and does not depend upon the nature of the liquid. So, in the given problem, the frequency would not be changed. 35. Why two organ pipes of same length open at both ends produce sounds of different frequencies if their radii are different? Ans.  [Here e = 0.6 r] Due to different radii, the frequencies are different. 36. Sound waves travel through longer distances during night than during day. Why? Ans. At night, the Earth’s atmosphere is warmer as compared to the surface of the Earth. The temperature increases with altitude. The velocity of sound increases. It is a case of reflection from denser to rarer medium. The sound waves get totally internally reflected. 37. Two tuning forks when sounded together give 3 beats per second. One is in unison with a length of 58 cm of a monochord string under constant tension and the other with 59 cm length of the same string. What are the frequencies of the forks? Ans. Suppose frequency of one tuning fork is n. Corresponding length of the wire, l = 58 cm, nl = constant   n = 177 Hz  Frequencies of forks are 174 Hz, 177 Hz. 38. When the wire of a sonometer is 73 cm long, it is in tune with a tuning fork. On shortening the wire by 5 mm, it makes 3 beats a second with the fork. What is the frequency of the fork? Ans. n  ,(n + 3)  Dividing, n = 435 Hz. 39. The audible frequency range of a human ear is 20 Hz – 20 kHz. Convert this into the corresponding wavelength range. Take the speed of sound in air at ordinary temperature to be 340 m s–1. Ans. Lower limit of wavelength,     or     = 17 mm  Upper limit of wavelength,    = 17 m  

342 Physics—XI 40. Find the speed of transverse waves in a copper wire having a cross sectional area of 1 mm2 under the tension produced by 1 kg wt. The relative density of copper = 8.93. Ans. a = 1 mm2 = 10–6 m2,  = 8.93 × 103 kg m–3 T = 1 kg wt = 9.8 N, mass/length,   = 10–6 × 1 × 8.93 × 103 kg m–1 = 8.93 × 10–3 kg m–1 v=  =   m s–1 = 33.13 m s–1 41. Calculate the speed of sound in oxygen from the following data. The mass of 22.4 litre of oxygen at STP (T = 273 K and P = 1.0 × 105 N m–2) is 32 g, the molar heat capacity of oxygen at constant volume is Cv = 2.5 R and that at constant pressure is Cp = 3.5 R. Ans.             =  = 1.4, P = 1.0 × 105 N m–2,  kg m–3  =   kg m–3 =    m s–1 = 313 m s–1  v= 42. Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 × 10–3 kg. Ans. We know that 1 mole of any gas occupies 22.4 litre at STP.  Density of air at STP,       For air,  Using Laplace’s formula   we get     = 331.1 m s–1 43. The velocity of sound in air at 47°C is 360 m s–1. Calculate the velocity of sound in air at 17°C. Ans. v47 = 360 m s–1, v17 =? ; Now     

Waves 343 or v17 =   = 342.7 m s–1 44. Calculate the velocity of longitudinal waves in hydrogen at NTP. Ans. We know that  = 1.4, R = 8.3 J mol–1 K–1, T = 273 K, M = 2 × 10–3 kg/mol Using,  we get v=   = 1259.42 m s–1 v=   45. At what temperature will the velocity of sound in hydrogen be twice as much as that at 27°C? Ans. =  4=  or   =  or or 273 + t = 1200 or t = 927°C 46. A sound wave of frequency 400 Hz is travelling in air at a speed of 320 m s–1. Calculate the difference in phase between two points on the wave 0.2 m apart in the direction of travel. Ans. The wavelength of the sound wave is   At 0.2 m apart, the phase difference is given by      = π rad   2 47. Given : = 0.8 sin 16π  +  metre. Calculate the wavelength and the velocity of  40  the wave represented by this equation. Ans. Rewriting the given equation,               Comparing with          or  = 8 Hz,  = 5 m Velocity, v =  = 40 m s–1 48. A sonometer wire carries a brass weight (specific gravity = 8) at its end and has a fundamental frequency of 320 Hz. What would be its frequency if this weight is completely immersed in water?

344 Physics—XI Ans. When the weight is immersed in water, buoyancy is , where T is the tension in the wire. Net tension = T –   = 320 Hz = 293.3 Hz 49. A wire is under tension of 32 N and length between the two bridges is 1 m. A 10 m length of the sample of the wire has a mass of 2 g. Deduce the speed of transverse waves on the wire and frequency of the fundamental. Ans. T = 32 N ;     ;L=1m    = 400 m s–1  Also,   = 200 Hz  50. Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. Speed of sound in air is 340 m s–1. Ans. n = n ×  Hz = n × 850 Hz  Fundamental frequency = 1 × 850 Hz = 850 Hz Frequency of first overtone = 2 × 850 Hz = 1700 Hz Frequency of second overtone = 3 × 850 Hz = 2550 Hz 51. An open organ pipe emits its fundamental note. Find its length if the velocity of sound in air is 330 m s–1 and it vibrates in unison with a string of length 35 cm under a stretching force of 2 kg wt. (Mass per unit length of the string is 0.04 g cm –1). Ans.   =   Hz = 100 Hz   Again,  or  = 2L Now, v =  = 2L or  = 1.65 m  52. A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. Speed of sound in air = 340 m s–1. Ans.  = or L=  m = 0.17 m = 17 cm  53. An open organ pipe emits a note of frequency 256 Hz which is its fundamental. What would be the smallest frequency produced by a closed pipe of the same length?

Waves 345 Ans. For open organ pipe,    For closed organ pipe,    = 128 Hz 54. The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe 2 m long. Calculate the length of the open pipe. Ans. Let Lo and Lc represent the lengths of open organ pipe and closed organ pipe respectively. Frequency of second overtone of open pipe = Frequency of first overtone of closed pipe = Equating,  or  or    = 4 m 55. A resonance air column resonates with tuning fork of frequency 512 Hz, at the length 17.4 cm. Neglecting the end correction, find the speed of sound in air. Ans.           = 356.4 m s–1 56. In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the air column has a length of 20.0 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find the speed of sound in air. (b) How much distance above the open end does the pressure node form? Ans. (a) v = 2 (l2 – l1) = 2 × 400 (62 – 20) × 10–2 m s–1 = 336 m s–1    cm = 1 cm (b) e = 57. In an experiment, it was observed that a tuning fork and a sonometer wire gave 5 beats per second both when the length of wire was 1 m and 1.05 m. Calculate the frequency of the fork. Ans. Let the frequency of the fork be . At the smaller length of the sonometer wire (l1 = 1 m), the frequency of the wire must be higher i.e.,     ; and at the larger length (l2 = 1.05 m), the frequency must be lower.    According to the law of length,       On solving, we get  = 205 Hz

346 Physics—XI 58. Two tuning forks A and B when sounded together give 4 beats /s. A is in unison with the note emitted by a 0.96 m length of a sonometer wire under a certain tension. B is in unison with 0.97 m length of the same wire under the same tension. Calculate the frequencies of the forks. Ans. A is in unison with a smaller length of the wire as compared to B. So, A has higher frequency as compared to B. Let  be the frequency of A. Then,           or     or   = 388 Hz 59. y1 = y0 sin [t – kx], y2 = y0 sin [t + kx]. Two waves when they are superimposed, we get progressive or standing wave? In terms of given data, state the velocity of y1 and y2 .  Ans. Velocity of y1 = in + ve x-direction.  Velocity of y2 = in – ve x-direction. The velocities of two waves are equal and opposite. Since the waves are travelling in opposite directions therefore on superimposition of two waves, standing wave is formed. 60. Two waves have equations : y1 = a sin (t + 1); y2 = a sin (t + 2). If the amplitude of the resultant wave is equal to the amplitude of each of the superposing waves, then what will be the phase difference between them? Ans. A2 = a2 + a2 + 2a2 cos (1 – 2) or a2 = 2a2 [1 + cos (1 – 2)] or cos (1 – 2) = – ; 1 – 2 = 120° =  radian. 61. Find the minimum stress to produce 1% strain, for string of density 4 × 103 kg m–3 and velocity of sound 5000 m s–1. Ans. v =  ; T = v2 ; F = 1% of T. 62. A transverse mechanical harmonic wave is travelling on a string. Maximum velocity and maximum acceleration of a particle on the string are 3 m s–1 and 90 m s–2 respectively. If wave is travelling with a speed 20 m s–1 on the string, write wave function describing the wave. Ans. A = 3, A2 = 90,   = 30 rad s–1 and A = 0.1 m  = 20  k = 1.5 The equation of wave will be y = A sin (t ± kx) or y = (0.1) sin (30t ± 1.5x).

Waves 347 63. Two identical sinusoidal waves, moving in the same direction along a stretched string, interfere with each other. The amplitude of each wave is 10.0 mm and the phase difference between them is 80°. What is the amplitude of the resultant wave and the nature of interference? Ans. Since the two waves are identical, they have same amplitudes. The amplitude of the resultant wave ym is given by : ym =  = 2 (10.0 mm) cos (80°/2) = 2 × 10 × 0.766 mm = 15.32 mm As the resultant amplitude is between 0 and 2ym, the interference is intermediate. 64. What is an echo and how is it produced? What should be the minimum distance of an obstacle so that an echo may be heard of (i) sharp momentary sound (ii) mono-syllabic sound? Given : speed of sound is 330 m s–1. Ans. (i) 16.5 m (ii) 33 m Hints. (i)  ; (ii)  65. A tuning fork produces resonance in a closed pipe. But the same tuning fork is unable to produce resonance in an open pipe of the same length. Why? Ans. Due to end-correction, the fundamental frequency of the open organ pipe is not exactly twice the fundamental frequency of the closed organ pipe. 66. A sonometer wire resonates with a tuning fork. If the length of the wire between the bridges is made twice, even then it can resonate with the same fork. Why? Ans. When the length of the wire is doubled, the fundamental frequency is halved. But now the frequency of the first harmonic will be n. So, the sonometer wire shall still resonate with the given tuning fork. But, now, the wire shall vibrate in two segments. 67. Why the speed of sound in moist air is greater than in dry air? Ans. This is because the density of water vapours is less than the density of dry air. Since  therefore the velocity of sound in moist air is greater than the velocity of sound in dry air. 68. Will the velocity of sound in moist hydrogen be greater than the velocity of sound in dry hydrogen? Ans. The density of moist hydrogen is greater than the density of dry hydrogen because the density of water vapour is greater than hydrogen. Since  therefore the velocity of sound in moist air will be less than the velocity of sound in dry hydrogen. 69. The disc of a siren having 60 holes makes 420 revolutions per minute. What is the frequency of sound produced? Given : velocity of sound = 100 m s –1. Ans.    

348 Physics—XI The product of number of holes and the number of revolutions per second gives the frequency of sound produced. 70. You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x – v  t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave: (a) (x – vt)2 ; (b) log [(x + vt)/x0 ] (c) 1/(x + vt) Ans. The converse is not true. An obvious requirement for an acceptable function for a travelling wave is that it should be finite everywhere and at all times. These functions are not finite for all values of x and t and hence they cannot represent a travelling wave. 71. A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound (b) the transmitted sound? Speed of sound in air is 340 m s–1, and in water is 1486 m s–1. Ans.  = 1000 × 103 Hz = 106 Hz, va = 340 m s–1, vw = 1486 m s–1 Wavelength of reflected sound, a =  = m = 3.4 × 10–4 m Wavelength of transmitted sound, w =  m = 1.486 × 10–3 m.  72. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1? The operating frequency of the scanner is 4.2 MHz. Ans.  =?, v = 1.7 km s–1 = 1700 m s–1  = 4.2 × 106 Hz ; v =  or  =  or  = m = 4.05 × 10–4 m = 0.405 mm.  73. A steel wire has a length of 12 m and a mass of 2.10 kg. What will be the speed of a transverse wave on this wire when a tension of 2.06 × 104 N is applied? Ans. v =   = 3.43 × 102 m s–1 74. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz? (sound velocity in air = 330 m s–1). Ans.  = (2n – 1)     , (2n – 1) =  , n=2 So, 2nd harmonic is resonantly excited by the given source. 75. The wave pattern on a stretched string is shown in Fig. 7.6. Interpret what kind of wave this is and find its wavelength.

Waves 349 t=0 10 20 30 50 x 0 x Displacement x x t = T/4 t = T/2 t = 3T/4 t=T x Fig. 7.6 Ans. The wave pattern represents a stationary wave. At t = 0, T, points at x = 0, 10, 20, 30, 40 are permanently at rest. These represent nodes.  Distance between successive nodes = = 10 cm or  = 20 cm SHORT ANSWER TYPE–II QUESTIONS (3 Marks) 1. The velocity of sound through hydrogen is 1400 m s–1. What will be the velocity of sound through a mixture of two parts by volume of hydrogen and one part of oxygen? Ans. If  be the density of hydrogen, then the density of oxygen is 16. If x be the total volume, then volume of hydrogen in mixture = , volume of oxygen in mixture  mass of hydrogen in mixture =  , mass of oxygen in mixture =    Total mass =  Density of mixture =    Now,  =

350 Physics—XI  or vm = vh     or vm =  = 571.4 m s–1. 2. For the plane waves, in air, of frequency 1000 Hz and displacement amplitude 0.2 × 10–7 m, deduce (i) the velocity amplitude (ii) the intensity. Given :  = 1.3 kg m–3, v = 340 m s–1. Ans.  = 1000 Hz, A = 0.2 × 10–7 m,  = 1.3 kg m–3, v = 340 m s–1 (i) velocity amplitude, v0 =          = 1.257 × 10–4 m s–1 (ii) intensity, I=    =     =     = 3.486 × 10–6 W m–2. 3. The fundamental frequency of a sonometer wire increases by 5 Hz if its tension is increased by 21%. How will the frequency be affected if its length is increased by 10%? Ans.    , where the letters have usual meanings.     Dividing,     or  or    If the length is increased by 10%, then     = 45.45 Hz.   4. Two tuning forks A and B when excited simultaneously produce 4 beats per second. The first fork A is in resonance with a closed organ pipe of length 16 cm while the second is in resonance with an open organ pipe of length 32.5 m. Find their frequencies. Ans. Frequency of closed organ pipe,     Frequency of open organ pipe,      Now,      or v = 4 × 64 × 65 cm s–1 Again,    = 260 Hz    = 256 Hz.

Waves 351 5. A tuning fork of frequency 340 Hz is allowed to vibrate just above a 120 cm high tube. Water is being filled slowly in the tube. What minimum height of water will be necessary for resonance? Speed of sound in air = 340 m s–1. Ans. We know that   or     For first resonance,   or  or  or l1 = 25 cm or  For second resonance,   For third resonance,   or   or  Clearly, the third resonance is not possible. Required minimum length of water column = 120 cm – maximum length of air column = 120 cm – 75 cm = 45 cm. 6. The intensities due to two sources of sound are I 0 and 4I0 . What is the intensity at a  point where the phase difference between two waves is (i) 0° (ii) (iii)  Ans. If a1 and a2 are the amplitudes of two waves, then the resultant amplitude is given by   , where  is the phase difference between two waves. Now,     Expressing this equation in terms of intensity,   (i)     (ii)     (iii)     . 7. An open pipe is suddenly closed at one end with the result that the frequency of the third harmonic of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. What is the fundamental frequency of the open pipe? Ans. Fundamental frequency of open pipe,  = Frequency of third harmonic of closed pipe,  = Now,  –  = 100 Hz