Title 15. Simple Harmonic Motion Med.pdf ✅

1. The equation of S.H.M. is y = a sin (2nt + ), then its phase at time is (a) 2nt (b)  (c) 2nt +  (d) 2t 2. The amplitude and the time period in a S.H.M. is 0.5 cm and 0.4 sec respectively. If the initial phase is /2 radian, then the equation of S.H.M. will be (a) y = 0.5 sin 5t (b) y = 0.5 sin 4t (c) y = 0.5 sin 2.5t (d) y = 0.5 cos 5t 3. Two equation of two S.H.M. are x = a sin (t – ) and y = b cos(t – ). The phase difference between the two is : (a) 0o (b) o (c) 90o (d) 180o 4. A particle is oscillating according to the equation X = 7 cos 0.5t, where t is in second. The point moves from the position of equilibrium to maximum displacement in time (a) 4.0 sec (b) 2.0 sec (c) 1.0 sec (d) 0.5 sec 5. A simple harmonic oscillator has an amplitude a and time period T. The time required by it to travel from x = a to x = a/2 is (a) T / 6 (b) T / 4 (c) T / 3 (d) T / 2 6. The amplitude and the periodic time of a S.H.M. are 5 cm and 6 sec respectively. At a distance of 2.5 cm away from the mean position, the phase will be (a) 5/12 (b) /4 (c) /3 (d) /6 7. A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement Y is (a) A 2 (b) A 2 (c) A 3 2 (d) 2A 3 8. A 1.00 x 10–20 kg particle is vibrating with simple harmonic motion with a period of 1.00 x 10–5 sec and a maximum speed of 1.00 x 103 m/s. The maximum displacement of the particle is : [ (a) 1.59 mm (b) 1.00 m (c) 10 m (d) None of these 9. A body is executing simple harmonic motion with an angular frequency 2rad/s. The velocity of the body at 20 mm displacement, when the amplitude of motion is 60 mm, is: (a) 40 mm / s (b) 60 mm / s (c) 113 mm / s (d) 120 mm / s 10. A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/sec. Its velocity will be 50 cm/sec at a distance (a) 5 (b) 5 2 (c) 5 3 (d) 10 2 11. A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The magnitude of the velocity in m sec–1 at the centre of oscillation is (a) 20  (b) 100 (c) 40  (d) 100 12. A particle executes S.H.M. with a period of 6 second and amplitude of 3 cm. Its maximum speed in cm/sec is (a)  / 2 (b)  (c) 2 (d) 3 13. A S.H.M. has amplitude ‘a’ and time period T. The maximum velocity will be (a) 4a T (b) 2a T (c) a 2 T  (d) 2 a T  14. A body is executing S.H.M. When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocity of the body is 10 cm/sec and 8 cm/sec. Then the time period of the body is (a) 2 sec (b)  / 2 sec (c)  sec (d) 3 / 2 sec
15. Which of the following is a necessary and sufficient condition for S.H.M. (a) Constant period (b) Constant acceleration (c) Proportionality between acceleration and displacement from equilibrium position (d) Proportionality between restoring force and displacement from equilibrium position 16. The acceleration of a particle in S.H.M. is (a) Always zero (b) Always constant (c) Maximum at the extreme position (d) Maximum at the equilibrium position 17. The displacement of a particle moving in S.H.M. at any instant is given by y = a sin t. The acceleration after time t = T 4 is (where T is the time period) (a) a (b) –a (c) a2 (d) –a2 18. The velocity of simple pendulum is maximum at : (a) extremes (b) half displacement (c) mean position (d) every where 19. A body is executing SHM. which of the following properties remains same at each point of its path during motion ? (a) Acceleration (b) velocity (c) phase (d) Total energy 20. The restoring force of SHM is maximum when particle: (a) displacement is maximum (b) half way between them (c) when crossing mean position (d) at rest 21. Amplitude of a pendulum is 60 mm and angular velocity is 2 rad s–1 . Find its velocity if its displacement is 22mm : (a) 120 mm s–1 (b) 112 mm s–1 (c) 130 mm s–1 (d) 125 mm s–1 22. A particle is executing the motion x = a cos (   t– ). The maximum velocity of the particle is : (a) a cos (b) a (c) a sin (d) none of these 23. A particle executing simple harmonic motion has a time period of 4s. After how much interval of time from t = 0 will its displacement be half of its amplitude ? (a) 1 s 3 (b) 1 2 s (c) 2 3 s (d) 1 6 s 24. The displacement x as a function of time t of a simple harmonic motion is represented by where A is a positive constant. (a) 2 2 d x dt – A2x = 0 (b) dx dt + A2x = 0 (c) 2 2 d x dt + A2x 2 = 0 (d) 2 2 d x dt + A2x = 0 25. The displacement y of a particle varies with time t, in second, as y = 2cos(t + /6). The time period of the oscillations is (a) 2 sec (b) 4 sec (c) 1 sec (d) 0.5 sec 26. If the acceleration-displacement graph of simple harmonic motion of a particle is given, then the time period of the particle will be
(a) 2 (b) 3 (c) 4 (d) 5 27. A student says that he had applied a force F = – k x on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he has worked only with positive x and no other force acted on the particle. (a) As x increases k increases (b) As x increases k decreases (c) As x increases k remains constant. (d) The motion cannot be simple harmonic. 28. For a particle executing simple harmonic motion, the acceleration is proportional to (a) displacement from the mean position (b) distance from the mean position (c) distance travelled since t = 0 (d) speed 29. The displacement of a particle in simple harmonic motion in one time period is (a) A (b) 2A (c) 4A (d) zero 30. The distance moved by a particle in simple harmonic motion in one time period is (a) A (b) 2A (c) 4A (d) zero 31. The time period of a particle in simple harmonic motion is equal to the time between consecutive appearance of the particle at a particular point in its motion. This point is (a) the mean position (b) an extreme position (c) between the mean position and the positive extreme (d) between the mean position and the negative extreme. 32. The average acceleration in one time period in a simple harmonic motion is (a) A 2 (b) A 2 /2 (c) A 2 / 2 (d) zero 33. A particle moves on the X-axis according to the equation x = A + B sin t. The motion is simple harmonic with amplitude (a) A (b) B (c) A + B (d) 2 2 A B+ 34. Equation of SHM is x = 10 sin 10t. Find the distance between the two points where speed is 50 cm/sec. x is in cm and t is in seconds. (a) 10 cm (b) 20 cm (c) 17.32 cm (d) 8.66 cm. 35. A mass M is performing linear simple harmonic motion, then correct graph for acceleration a and corresponding linear velocity  is (a) (b) (c) (d)
36. A particle moves on the X-axis according to the equation x = x0 sin2 t. The motion is simple harnomic (a) with amplitude x0 (b) with amplitude 2x0 (c) with time period 2  (d)with time period   37. The motion of a particle is given by x = A sin t + B cos t. The motion of the particle is (a) not simple harmonic (b) simple harmonic with amplitude A + B (c) simple harmonic with amplitude (A + B)/2 (d) simple harmonic with amplitude 2 2 A B+ 38. A simple harmonic motion is given by y = 5 (sin 3t + 3 cos 3 t). What is the amplitude of motion if y is in m ? (a) 100 cm (b) 5 m (c) 200 cm (d) 1000 cm 39. The position vector of a particle from origin is given by r = A ( ˆ i cost + ˆ j sint). The motion of the particle is (a) simple harmonic (b) on a straight line (c) on a circle (d) with constant acceleration 40. The amplitude of a particle executing S.H.M with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is - (a) 2 2 144 m/ sec  (b) 2 144m/ sec (c) 2 144 2 m/ sec  (d) 2 2 288 m/ sec  41. At how many maximum points, velocity of simple pendulum is same : (a) 1 (b) 2 (c) 4 (d) 6 42. Velocity at mean position of a particle executing S.H.M. is , then velocity of the particle at a distance equal to half of the amplitude : (a) 4v (b) 2 (c) 3 2  (d) 3 4  43. A particle is executing S.H.M. of frequency 300 Hz and with amplitude 0.1 cm, Its maximum velocity will be : (a) 60  cm/s (b) 0.6  cm/s (c) 0.50  cm/s (d) 0.05  cm/s 44. The velocity of a particle in S.H.M. at displacement y from mean position is (a = amplitude,  = angular frequency) (a) 2 2  + a y (b) 2 2  a – y (c) y (d) 2 (a2 – y 2 ) 45. The displacement of a particle varies according to the relation x = 4 (cos t + sin t). The amplitude of the particle is : (a) –4 (b) 4 (c) 4 2 (d) 8 46. A particle executes simple harmonic motion between X = – A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2 , then (a) T1 < T2 (b) T1 > T2 (c) T1 = T2 (d) T1 = 2 T2 47. The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4/3 s is (a) 3 2 32  cm/s2 (b) 2 32 − cm/s2 (c) 2 32  cm/s2 (d) – 3 2 32  cm/s2 0 4 8 12 t(s) 1 –1 x(cm)