SOLUTIONS TO SHM & Wave-PART TEST - 1

SOLUTIONS TO PART TEST - 1 (PT-1) TOPIC : SHM & WAVE 1. From the given information it can be concluded that points A and B are equidistant from mean position. Hence from diagram it is clear that time period of oscillation is = 2 + 2 × 2 + 2 = 8 second. 2. Maximum velocity v =  a Maximum acceleration f = 2 a  f = v a 3. When the particle crosses D, its speed is half the maximum speed.  v = or vmax 2 vmax R  R or x = 2 R  Distance BD = 2x = 3 R K(2K) 2K 4. Both the spring are in series  K = K  2K = 3 Time period T = 2 where  = m1m2 m1  m2 Here μ = m  T = 2π 2 = 2 Method II ∴ mx = mx ⇒ x = x force equation for first block; 2k 3 (x1 + x2 ) = –m d2 x1 dt2 Put x = x  d2 x1 dt2 4k + 3m × x1 = 0 4k  2 = 3m  T = 2 5. Resulting torque on the bob = mg 𝑙 sin MI of bob about axis xy = For small angle  m𝑙2 2   =  = 2 g  `  = 𝑙  T = 2 mg 1 6. f0 = 2 where, 𝑙 is distance between point of suspension and centre of mass of the body. Thus, for the stick of length L and mass m : 1 f0 = 2 1 = 2 when bottom half of the stick is cut off 1 f0’ = 2 = 1  2 = f0 7.  = – k 0.1 = – k(1.0), where k is torsional constant of the wire. 1 k = 10 T = 2 = 2 = 2 10  0.2  0.2 10 = 4 second Ans. d 8. dt = 2   = 2t y Let BP = a  x = OM = a sin = a sin(2t) Hence M executes SHM within the given time period and its acceleration is opposite to ‘x’ that means towards left x 9. Average kinetic energy with respect to position = 1 kA 2 3 Average potential energy with respect to position = 1 kA 2 6 Average kinetic energy with respect to time = 1 kA2 4 Average potential energy with respect to time = 10. After the system is released, m2 moves down. The extention in the spring becomes : 1 kA2 4 m2g (m g = kx ), which is the new equilibrium position of the system. k 2 0 For small 'x' : restoring force on the system is F = kx  a = kx m1  m2 (For (m1 + m2 + spring) system)  T = 2 = 2 = 2  Angular frequency =  = 2 = . T F.B.D. of m1 and m2 just after the system is released : From above : T = m1m2 g 1 2 Hence (C) is incorrect. After x = m2g ; m k 1 moves towards right till the total kinetic energy acquired does not converted to potential energy. Hence (D) is also incorrect. Hence (B) is the answer. 11. P.E. is maximum at extreme position and minimum at mean position. Time to go from extreme position to mean position is, t = T ; where T is time period of SHM 4 5 s = T  T = 20 s. 4 12. E1 = 1 kx2 and E = 2 1 2 ky2 Therefore E = 1 k(x + y)2 = 2 1 kx2 + 2 1 ky2 + kxy = E 2 1 + E2 + k  = E1 + E2 + 2 13. Speed of block is maximum at mean position. At mean position upper spring is extended and lower spring is compressed. 14. 1 mV 2 = 15 × 10–3  V = m / s 2 m A = m / s m  L  . = m / s 0.150 = 100 103  L = 0.150 0.1 = 1.5 m 15. The magnitude of phase difference between the points separated by distance 10 metres = k × 10 = [10π × 0.01] × 10 = π 16. V max = A = 5 cm/s ; T = 4 sec   = 2 =  4 2  A = 5  / 2 10 =  cm.   17. v =  2  2  1 2  r        =  2  2𝑙1 2𝑙 2  r   1  2𝑙1 × 1 = 2 1 2𝑙 2    𝑙 : 𝑙 2 = 1 : 2 18. For waves along a string : v     Now, for 6 loops : 3 = L 1   = L/3 1 & for 4 loops : 2 = L 2   = L/2 2  1  2  T = 9 xT = 9 × 36 = 81 N. 2 3 2 4 1 4 1 19. P = 2 2A2V using V = 1  P = 2 2 A2  =  f =  2 1 = 2 using data f = 30 Hz. 20. In Sonometer f  V  (Wavelength constant for fixed length) V1  V2 = 2 = T  T2 = 4  % T  T T  T1 1 2 T1 × 100 = 4 × 100 = 75% T1 21. P0 = B.K.S0 =  2     0  P  1 0    Thus, pressure amplitude is highest for minimum wavelength, other parameters B and S0 being same for all. From given graphs.  <  <  . Hence (B). 22. The speed of sound in air is v =  M of H2 is maximum, hence speed of sound in H2 shall be maximum. 23.   P 2 ' P0 '2  = P0  P ’ = P  P  = P 24. For pipe A, second resonant frequency is third harmonic thus f = 3V 4LA For pipe B, second resonant frequency is second harmonic thus f = 2V 2LB Equating, 3V 4LA 2V = 2LB  L = 4 3 LA = 4 3 .(1.5) = 2m. 25. For a stationary observer between wall and source,  v  frequency from direct source =  v  v  f0 s   v  frequency from reflected sound =  v  v  f0 . s  So no beats will be heard.  v   v  26.  v  v    v  v  f0 = 2 Hz  v = 0.5 m/s  s   s  27. After a time t, velocity of observer V0 = at  V  V0   V  at   f =   f . =   f , which is a straight line graph of positive slope. 0  V  s  V  s 28. Beats heard per second by A is 8 and B is zero. 29. Phase of the motion is (t + ). Using x = A sin (t + ) and V = A cos(t + ) for conditions at t = 0  x = A and V = 0 then  = /2 When it passes equilibrium position for the first time t = T 4 Phase = 2. T   =  T 4 2 30.  = 384 – 288 = 96 Thus 288 and 384 (96 × 3; 96 × 4) are third and fourth harmonics. For fundamental mode:  2 = 0.75  = 1.5 m  = 96  v = 96 × 1.5 = 144 m/s. SOLUTIONS TO PART TEST - 2 (PT-2) TOPIC : SHM & WAVE 1. The magnitude of displacement in the given time interval = a 2 Time taken by the particle to cover a distance a starting from rest = T 2 6 a / 2 3a Hence the magnitude of average velocity over given time interval is = T / 6 = T 2. With respect to the cart, equilibrium position of the pendulum is shown. If displaced by small angle  from this position, then it will execute SHM about this equilibrium position, time period of which is given by : T = 2  ; geff = g = 2g  T = 1.0 second 3. x = 3 sin 100 t y = 4 sin 100 πt Equation of path is y  4 x 3 i.e. y  4 x 3 4 which is equation of a straight line having slope = 3 Equation of resulting motion is → r  xi  yj = (3ˆi  4ˆj) sin 100t Amplitude is = 5 4. Reflected pulse will be inverted as it is reflected by a denser medium. The wall exerts force in downward direction. 5. The equation of pressure variation due to sound is  s p = – B  x = – B   x [s0 sin2 (t – kx)] = B ks0 sin (2t – 2kx) P2 6. Ι = 0 ⇒ P 4r 2 2 0 2V ; where P, P0 , V are power, pressure amplitude and velocity respectively. ⇒ P = = = 5  7. L = (2n – 1) 4 4L  = (2n – 1) 1 = 4L 2(1)  1 = 4L 2 = 4L 4L 4  1 = 3 3 = 4L 4L 6 – 1 = 5 8. Two consecutive frequencies are 420 Hz & 480 Hz. So the fundamental frequency will be 60 Hz.  60 = 1 2  𝑙  𝑙 = 2.5 m 9. At time t1, velocity of the particle is negative i.e. going towards –Xm. From the graph, at time t1, its speed is decreasing. Therefore particle lies in between –Xm and 0. At time t2, velocity is positive and its magnitude is less than maximum i.e. it has yet not crossed O. It lies in between –Xm and 0. Phase of particle at time t1 is (180 + 1). Phase of particle at time t2 is (270 + 2) Phase difference is 90 + (2 – 1) 2 – 1 can be negative making  < 90° but can not be more than 90°. 10. y = 2A sin kx. sint dy Vy = dt = 2A sin kx. cos  t T  2  V = 0  t = T/4, 3T/4     (2 times in one time period)  11. 4 = 0.1   = 0.4 m from graph  T = 0.2 sec. and amplitude of standing wave is 2A = 4 cm. Equation of the standing wave  2   2  y(x, t) = – 2A cos  0.4 x . sin  0.2 t  cm     y (x = 0.05, t = 0.05) = – 2 cm  y(x = 0.04, t = 0.025) = – 2 cos 36°  y  2 1 . cos 2   speed = = 2 m/sec.  Vt = t = – 2A 10  cos  0.4 15   0.1 0.2 = 20  cm/s T     12. The speed of the particle (x, t) is  y    t 12 (x  3t) [(x  3t)2  1]2 . Therefore at x = 3 and t = 1,  y  0  t The speed of the pulse = coefficient of t coefficient of x = 3.0 cm/s 13. The mean position of the particle in statement-1 is x =  b a and the force is always proportional to displacement from this mean position. The particle executes SHM about this mean position. Hence statement-1 is false 14. Two waves moving in uniform string with uniform tension shall have same speed and may be moving in opposite directions. Hence both waves may have velocities in opposite direction. Hence statement-1 is false. 15. Every small segment is acted upon by forces from both sides of it hence energy is not conserved, rather it is transmitted by the element. 16. The principle of superposition of waves is always valid. 17. When speed of block is maximum, net force on block is zero. Hence at that instant spring exerts a force of magnitude 'mg' on block. 18. At the instant block is in equilibrium position, its speed is maximum and compression in spring is x given by kx = mg (1) From conservation of energy mg (L + x) = 1 kx2 + 2 1 mv2 2 max .... (2) from (1) and (2) we get v max = . 19. V max = and  = = 2  A = Vmax  = 3 L 4 Hence time taken t, from start of compression till block reaches mean position is given by x = A sin  t where x = L 4  t = 1 sin–1 3 2 Time taken by block to reach from mean position to bottom most position is 4 = Hence the required time = 1 + sin–1 3 20. Difference in path travelled by the sound = hd D hence time gap between the recordings = hd Dv 21. For resultant pressure amplitude to be maximum, path difference. D  m hd  m v   mvD  D f f hd  For minimum frequency, m = 1   vD f hd  22. If the source is shifted slightly towards positive x-direction (say x). v(D  x) Then minimum frequency will be = hd hence minimum frequency will decrease 23. (A) In frame of lift effective acceleration due to gravity is g  g  3g downwards 2 2  T = 2 (B) K𝑙 = mg  k  g m L constant acceleration of lift has no effect in time period of oscillation.  T = 2 = 2 (C) T = 2 = 2  2 (D) T = 2 = 2 = 2 24. (A) Number of loops (of length /2) will be even or odd and node or antinode will respectively be formed at the middle. Phase of difference between two particle in same loop will be zero and that between two particles in adjacent loops will be . (B) and (D) Number of loops will not be integral. Hence neither a node nor an antinode will be formed in in the middle. Phase of difference between two particle in same loop will be zero and that between two particles in adjacent loops will be . (C) Number of loops (of length /2) will be even or odd and antinode or node will respectively be formed at the middle. Phase of difference between two particle in same loop will be zero and that between two particles in adjacent loops will be . SOLUTIONS TO PART TEST - 3 (PT-3) TOPIC : SHM & WAVE 2 1. x = A sin T t ; for x = A 2 A 2  2 = A sin T t T 1 Solving t = 12 = 6 second. 2. Restoring torque :  = ky L = KL2 (Since y  L from figure)  kL2 = mL2 3 .    = 3k · m   = (torque due to mg was already balanced so it is not taken in calculation) 3. As wave has been reflected from a rarer medium, therefore there is no change in phase. Hence equation for the opposite direction can be written as y = 0.5A sin (–kx – t + ) = – 0.5A sin (kx + t – ) 4.  = 2𝑙 = 3m Equation of standing wave y = 2A sin kx cos t y = A as amplitude is 2A. A = 2A sin kx 1  x = 4 m and 5. V =  x  B = V2 = 1.25 m  x – x1 = 1m = (5.40 × 103 m/s)2 (2.7 × 103) = 7.9 × 1010 Pa. 3V 6. Second overtone of open pipe = 2𝑙1 5V second overtone of closed pipe = 4𝑙 2 Since, ratio of frequency are same  3V  5V  𝑙1  4  3 = 6 2𝑙1 4𝑙 2 𝑙 2 2  5 5 Now, the ratio of fundamental frequencies : V 1  2𝑙1  2𝑙 2 = 10 : 6 = 5 : 3  2 V 4𝑙 2 𝑙1 7. When the point of suspension is at a distance x from centre of length of rod, the time period of oscillation is T = 2p where l is length of the rod. The time period of oscillation will be same (T ) if the point of suspension is a distance x = /2 or x = /6 from centre of the rod. Thus there will be three additional points. 8. At points A and B, path difference between the waves coming from two B radio stations is zero. Hence there will be constructive interference at A and B, s s2  For point C, path difference between the waves is 50 metre i.e. 2 so destructive interference takes places at point C. 9. At t = 0  Displacement x = x1 + x2 = 4 sin 3 Resulting Amplitude A = Maximum speed = A = 20 m/s = 2 m.   = 2 m Maximum acceleration = A2 = 200 m/s2 1 Energy of the motion = m2 A2 = 28 J 2 10. For standing wave to be formed the interfering waves must have same amplitude, same frequency and opposite direction of traveling. These are satisfied by options A, B and C. 11. T = 2  f = = all are independent of amplitude but Max. K.E. = 1 m2 a2  depends on amplitude. 2 V V 12. For first resonance with 400 Hz tuning fork 𝑙eq = 4f0 = 4(400) = (19 + 1) = 20 cm If we use 1600 Hz tuning fork for Resonance V 4f0  V 4  (1600)  20 4 = 5 cm 400 Hz 𝑙eq = V , 4f0 3V 4f0 , 5V , 4f0 7V , .... 4f 20 cm 0 1 cm + 𝑙 = 5 cm , 15 cm , 25 cm , 35 cm , 45 cm ..... 𝑙 = 4 cm , 14 cm , 24 cm , 34 cm , 44 cm ..... water level should be further lowered by 24 – 19 = 5 cm 34 – 19 = 15 cm 13. Statement-2 itself explains statement-1. 14. Speed of wave in a medium is property of the medium. 15. The equation of wave moving in negative x-direction, assuming origin of position at x = 2 and origin of time (i.e. initial time) at t = 1 sec. y = 0.1 sin (4t + 8x) Shifting the origin of position to left by 2m, that is, to x = 0. Also shifting the origin of time backwards by 1 sec, that is to t = 0 sec. y = 0.1 sin [(4(t–1) + 8(x – 2)] 16. As given the particle at x = 2 is at mean position at t = 1 sec.  its velocity v = A = 4 × 0.1 = 0.4  m/s. 17. Time period of oscillation T = 2  2  1 sec.  4 2 Hence at t = 1.125 sec, that is, at T seconds after t = 1 second, the particle is at rest at extreme position. 4 Hence instantaneous power at x = 2 at t = 1.125 sec is zero. v 2 18. The velocity-displacement relation is x2 + 2 = A2 which may be a circle if  = 1. and ellipse of   1. a2 The acceleration – velocity relation is v2 + 2 = A2 which maybe a circle if  = 1 and ellipse of   1. Acceleration-displacement graph is straight and acceleration time graph is sinusoidal. 19. The coordinates of the particles are x1 = A1 cos t, x = A2 cos t separation = x1 – x2 = (A1 – A ) cos t = 12 cos t Now x1 – x2 = 6 = 12 cos t  t =  3  2 12  . t = 3  t = 2s 20. The bob will execute SHM about a stationary axis passing through AB. If its effective length is 𝑙' then T = 2 𝑙' = 𝑙/ sin = 2 𝑙 (because  = 45) g' = g cos = g/ 2𝑙 T = 2 g = 2 2  0.2 10 2 = 5 s. X = 4 mg 21. Imagine a cylinder of radius 7m and length 10m. Intensity of sound at the surface of cylinder is same everywhere. P Therefore  = 2rL (As sound is propagating radially out only, sound energy does not flow out through the ends) P   = 50 W/m2 Energy intercepted by the detector =  × A = 12 mW 22.  Acceleration of the wave pulse remains constant in this case  distance = .t  CB = 𝑙 = VC = 0 , VB = VC  VB t 2 CB = and BA = 𝑙 = , VB´ = VB´  VA t . 2 BA = , VA = = , 𝑙 = 2 t , 𝑙 = 2g𝑙  CB 2 3g𝑙 tBA  2𝑙 2 𝑙  1  9610  1  t = tCB + tBA = = 1  2   = 2  168110 1 1.4  1.7  961  4.1 = 2 1681  3.1 = 2 = 2 Ans.