{"id":12641,"date":"2025-09-20T05:34:38","date_gmt":"2025-09-20T04:34:38","guid":{"rendered":"https:\/\/mcqsadda.com\/?p=12641"},"modified":"2025-10-22T09:45:54","modified_gmt":"2025-10-22T08:45:54","slug":"simple-harmonic-motion-top-100-mcqs-with-answer-and-explanation","status":"publish","type":"post","link":"https:\/\/mcqsadda.com\/index.php\/2025\/09\/20\/simple-harmonic-motion-top-100-mcqs-with-answer-and-explanation\/","title":{"rendered":"Simple harmonic motion Top 100 MCQs With Answer and Explanation"},"content":{"rendered":"\n

1. Simple harmonic motion is defined as:<\/mark><\/strong>
a) Motion with constant velocity
b) Motion with acceleration proportional to displacement and directed opposite to it
c) Motion with constant acceleration
d) Motion with displacement proportional to velocity
Answer<\/strong>: b
Explanation<\/strong>: SHM is a periodic motion where restoring force (or acceleration) is directly proportional to displacement and acts in the opposite direction, i.e., a\u221d-x.<\/p>\n\n\n\n

2. The displacement in SHM is given by:<\/mark><\/strong>
a) x=Asin\u2061(\u03c9t+\u03d5)
b) x=Acos\u2061(\u03c9t+\u03d5)
c) Both (a) and (b)
d) None of these
Answer<\/strong>: c
Explanation<\/strong>: Displacement in SHM can be represented by sine or cosine functions depending on initial conditions.<\/p>\n\n\n\n

3. The time period of a simple harmonic oscillator depends on:
<\/mark><\/strong>a) Amplitude only
b) Mass and restoring constant
c) Velocity of oscillation
d) None of these
Answer: b
Explanation: Time period T=2\u03c0\u221a(m\/k). It is independent of amplitude.<\/p>\n\n\n\n

4. In SHM, acceleration is maximum when:
<\/mark><\/strong>a) Displacement is zero
b) Displacement is maximum
c) Velocity is maximum
d) At equilibrium position
Answer:<\/strong> b
Explanation:<\/strong> a=-\u03c9^2 x. Maximum acceleration occurs at maximum displacement.<\/p>\n\n\n\n

5. Velocity in SHM is maximum when:
<\/mark><\/strong>a) Displacement is zero
b) Displacement is maximum
c) Acceleration is maximum
d) At turning points
Answer:<\/strong> a
Explanation:<\/strong> At equilibrium (x=0), velocity is maximum.<\/p>\n\n\n\n

6. The phase difference between velocity and displacement in SHM is:
<\/mark><\/strong>a) 0^\u2218
b) 90^\u2218
c) 180^\u2218
d) 270^\u2218
Answer:<\/strong> b
Explanation: <\/strong>Velocity leads displacement by \u03c0\/2in SHM.<\/p>\n\n\n\n

7. Which of the following is an example of SHM?
<\/mark><\/strong>a) Uniform circular motion projection
b) A pendulum oscillating with small amplitude
c) A mass attached to a spring
d) All of these
Answer:<\/strong> d
Explanation<\/strong>: All these are classic examples of SHM.<\/p>\n\n\n\n

8. In SHM, kinetic energy is maximum when:
<\/mark><\/strong>a) Displacement is maximum
b) Displacement is zero
c) Acceleration is maximum
d) None
Answer<\/strong>: b
Explanation:<\/strong> KE = maximum at equilibrium (displacement = 0).<\/p>\n\n\n\n

9. Which of the following is true about potential energy in SHM?
<\/mark><\/strong>a) Maximum at equilibrium
b) Minimum at extreme position
c) Maximum at extreme position
d) Constant everywhere
Answer:<\/strong> c
Explanation:<\/strong> Potential energy is maximum at maximum displacement.<\/p>\n\n\n\n

10. A pendulum performs SHM only if:
<\/mark><\/strong>a) Angle of oscillation is very small
b) Mass of bob is large
c) Length of string is small
d) For all cases
Answer: <\/strong>a
Explanation:<\/strong> For small angles, sin\u2061\u03b8\u2248\u03b8, making the motion approximately SHM.
11. The restoring force in SHM is proportional to:
<\/mark><\/strong>a) Velocity
b) Displacement
c) Acceleration
d) Momentum
Answer<\/strong>: b
Explanation<\/strong>: Restoring force F=-kx.<\/p>\n\n\n\n

12. If a body in SHM has amplitude A, the maximum velocity is:
<\/mark><\/strong>a) \u03c9A
b) A\/\u03c9
c) A\u03c9^2
d) None
Answer<\/strong>: a
Explanation<\/strong>: v_”max” =\u03c9A.<\/p>\n\n\n\n

13. The total energy in SHM is proportional to:
<\/mark><\/strong>a) Amplitude
b) Amplitude squared
c) Velocity
d) Frequency
Answer<\/strong>: b
Explanation<\/strong>: E=1\/2 kA^2, proportional to A^2.<\/p>\n\n\n\n

14. The frequency of SHM is:
<\/mark><\/strong>a) Independent of amplitude
b) Directly proportional to amplitude
c) Inversely proportional to amplitude
d) Depends on energy
Answer<\/strong>: a
Explanation<\/strong>: Frequency depends on system parameters, not amplitude.<\/p>\n\n\n\n

15. The graph of displacement vs time in SHM is:
<\/mark><\/strong>a) Straight line
b) Exponential
c) Sinusoidal
d) Parabola
Answer<\/strong>: c
Explanation:<\/strong> Displacement is a sine or cosine function of time.<\/p>\n\n\n\n

16. In SHM, velocity is zero when:
<\/mark><\/strong>a) Acceleration is maximum
b) Displacement is zero
c) Energy is maximum
d) At equilibrium
Answer<\/strong>: a
Explanation<\/strong>: At extreme positions, acceleration is maximum, velocity is zero.<\/p>\n\n\n\n

17. The time period of a simple pendulum is independent of:
<\/mark><\/strong>a) Length of pendulum
b) Value of acceleration due to gravity
c) Mass of bob
d) Amplitude (small)
Answer:<\/strong> c
Explanation:<\/strong> T=2\u03c0\u221a(l\/g), mass has no effect.<\/p>\n\n\n\n

18. The dimension of angular frequency \u03c9is:
<\/mark><\/strong>a) [M^0 L^0 T^(-1)]
b) [M^0 L^1 T^(-2)]
c) [M^0 L^0 T^0]
d) [M^1 L^0 T^(-2)]
Answer:<\/strong> a
Explanation<\/strong>: Angular frequency has units of rad\/s, i.e. T^(-1).<\/p>\n\n\n\n

19. Which one is not an SHM?
<\/mark><\/strong>a) Motion of pendulum (small angle)
b) Motion of a particle on spring
c) Motion of earth around sun
d) Vibration of tuning fork
Answer:<\/strong> c
Explanation<\/strong>: Earth\u2019s motion is elliptical, not SHM.<\/p>\n\n\n\n

20. The frequency of oscillation is reciprocal of:
<\/mark><\/strong>a) Amplitude
b) Time period
c) Phase angle
d) Velocity
Answer:<\/strong> b
Explanation<\/strong>: Frequency = 1\/T.<\/p>\n\n\n\n

21. If the time period is doubled, the frequency becomes:
<\/mark><\/strong>a) Double
b) Half
c) Same
d) Four times
Answer<\/strong>: b
Explanation:<\/strong> Frequency = inverse of time period.<\/p>\n\n\n\n

22. In SHM, potential energy is zero when:
<\/mark><\/strong>a) Displacement is maximum
b) Displacement is zero
c) Velocity is zero
d) Acceleration is maximum
Answer:<\/strong> b
Explanation:<\/strong> At equilibrium, potential energy is zero.<\/p>\n\n\n\n

23. Which physical quantity remains constant in SHM?
<\/mark><\/strong>a) Velocity
b) Acceleration
c) Total energy
d) Potential energy
Answer:<\/strong> c
Explanation<\/strong>: Total energy remains constant (conservation of energy).<\/p>\n\n\n\n

24. SHM is a type of:
<\/mark><\/strong>a) Uniform motion
b) Periodic motion
c) Rectilinear motion
d) Projectile motion
Answer:<\/strong> b
Explanation<\/strong>: SHM is a periodic oscillatory motion.<\/p>\n\n\n\n

25. The graph between restoring force and displacement in SHM is:
<\/mark><\/strong>a) Straight line passing through origin
b) Parabola
c) Exponential
d) Constant
Answer:<\/strong> a
Explanation:<\/strong> F=-kx, a straight line graph with negative slope.
<\/p>\n\n\n\n

26. The general equation of displacement in SHM is:
<\/mark><\/strong>a) x=Asin\u2061(\u03c9t)
b) x=Acos\u2061(\u03c9t)
c) x=Asin\u2061(\u03c9t+\u03d5)
d) All of these
Answer<\/strong>: d
Explanation<\/strong>: Depending on initial conditions, SHM displacement can be expressed using sine, cosine, or with phase angle.<\/p>\n\n\n\n

27. If displacement is x=Asin\u2061(\u03c9t), then velocity is:
<\/mark><\/strong>a) v=A\u03c9cos\u2061(\u03c9t)
b) v=-A\u03c9sin\u2061(\u03c9t)
c) v=Acos\u2061(\u03c9t)
d) v=Asin\u2061(\u03c9t)
Answer<\/strong>: a
Explanation:<\/strong> Velocity is derivative of displacement: v=dx\/dt=A\u03c9cos\u2061(\u03c9t).<\/p>\n\n\n\n

28. The maximum acceleration in SHM is:
<\/mark><\/strong>a) A\u03c9
b) A\u03c9^2
c) \u03c9^2\/A
d) 1\/(A\u03c9^2)
Answer<\/strong>: b
Explanation:<\/strong> a=-\u03c9^2 x. Maximum acceleration = \u03c9^2 A.<\/p>\n\n\n\n

29. The phase difference between acceleration and displacement in SHM is:
<\/mark><\/strong>a) 0^\u2218
b) 90^\u2218
c) 180^\u2218
d) 270^\u2218
Answer:<\/strong> c
Explanation:<\/strong> Since a=-\u03c9^2 x, acceleration is always opposite to displacement (phase difference \u03c0).<\/p>\n\n\n\n

30. The kinetic energy of a particle in SHM is given by:
<\/mark><\/strong>a) 1\/2 kx^2
b) 1\/2 m\u03c9^2 (A^2-x^2)
c) 1\/2 mv^2
d) Both b and c
Answer<\/strong>: d
Explanation<\/strong>: KE in SHM can be expressed in terms of displacement or velocity.<\/p>\n\n\n\n

31. The potential energy of SHM is:
<\/mark>a) 1\/2 m\u03c9^2 x^2
b) 1\/2 mv^2
c) 1\/2 kv^2
d) 1\/2 A^2
Answer:<\/strong> a
Explanation<\/strong>: Potential energy = work done by restoring force = 1\/2 kx^2=1\/2 m\u03c9^2 x^2.<\/p>\n\n\n\n

32. The total energy in SHM is given by:
<\/mark><\/strong>a) 1\/2 m\u03c9^2 A^2
b) 1\/2 m\u03c9^2 x^2
c) 1\/2 mv^2
d) None
Answer:<\/strong> a
Explanation: <\/strong>Total mechanical energy = constant = 1\/2 m\u03c9^2 A^2.<\/p>\n\n\n\n

33. At equilibrium, in SHM:
<\/mark><\/strong>a) KE is maximum, PE is zero
b) KE is zero, PE is maximum
c) Both KE and PE are equal
d) Total energy is zero
Answer:<\/strong> a
Explanation<\/strong>: At equilibrium x=0, PE=0, KE = maximum.<\/p>\n\n\n\n

34. At extreme positions, in SHM:<\/mark><\/strong>
a) KE is maximum, PE is zero
b) KE is zero, PE is maximum
c) Both KE and PE are equal
d) KE = PE
Answer:<\/strong> b
Explanation<\/strong>: At extreme displacement, velocity = 0, so KE=0, PE=maximum.<\/p>\n\n\n\n

35. The ratio of kinetic energy to total energy in SHM is:
<\/mark><\/strong>a) x^2\/A^2
b) 1-x^2\/A^2
c) A^2\/x^2
d) Constant
Answer:<\/strong> b
Explanation<\/strong>: KE = 1\/2 m\u03c9^2 (A^2-x^2), total = 1\/2 m\u03c9^2 A^2. Ratio = 1-(x^2\/A^2).<\/p>\n\n\n\n

36. The ratio of potential energy to total energy in SHM is:
<\/mark><\/strong>a) x^2\/A^2
b) 1-x^2\/A^2
c) Constant
d) Zero
Answer:<\/strong> a
Explanation<\/strong>: PE = 1\/2 m\u03c9^2 x^2, total = 1\/2 m\u03c9^2 A^2. Ratio = x^2\/A^2.<\/p>\n\n\n\n

37. The frequency of SHM depends on:
<\/mark><\/strong>a) Amplitude
b) Mass and spring constant
c) Velocity
d) Energy
Answer<\/strong>: b
Explanation:<\/strong> For spring-mass system, f=1\/2\u03c0 \u221a(k\/m).<\/p>\n\n\n\n

38. For a pendulum of length l, the time period is:
<\/mark><\/strong>a) 2\u03c0\u221a(g\/l)
b) 2\u03c0\u221a(l\/g)
c) \u03c0\u221a(l\/g)
d) 2\u03c0 l\/g
Answer:<\/strong> b
Explanation<\/strong>: Time period of simple pendulum = T=2\u03c0\u221a(l\/g).<\/p>\n\n\n\n

39. In SHM, the restoring force is always directed:
<\/mark><\/strong>a) Along velocity
b) Opposite to velocity
c) Towards mean position
d) Away from mean position
Answer:<\/strong> c
Explanation<\/strong>: Restoring force always tries to bring particle back to equilibrium.<\/p>\n\n\n\n

40. If the amplitude is doubled, the maximum velocity becomes:
<\/mark><\/strong>a) Half
b) Double
c) Same
d) Four times
Answer:<\/strong> b
Explanation<\/strong>: v_max=\u03c9A, so directly proportional to amplitude.<\/p>\n\n\n\n

41. If amplitude is doubled, the total energy becomes:
<\/strong><\/mark>a) Double
b) Same
c) Four times
d) Half
Answer<\/strong>: c
Explanation<\/strong>: Total energy \u221d A^2. So doubling amplitude \u2192 4\u00d7 energy.<\/p>\n\n\n\n

42. At half the amplitude, the potential energy is:
<\/mark><\/strong>a) 1\/4of total energy
b) 1\/2of total energy
c) 3\/4of total energy
d) Equal to total energy
Answer:<\/strong> a
Explanation:<\/strong> U\/E=(x^2\/A^2)=(A\/2)^2\/A^2=1\/4.<\/p>\n\n\n\n

43. At half the amplitude, the kinetic energy is:
<\/mark><\/strong>a) 1\/4of total energy
b) 3\/4of total energy
c) Equal to potential energy
d) Zero
Answer:<\/strong> b
Explanation<\/strong>: KE\/E = 1 \u2013 (x^2\/A^2) = 1 \u2013 1\/4 = 3\/4.<\/p>\n\n\n\n

44. The graph of total energy vs time in SHM is:
<\/mark><\/strong>a) Sinusoidal
b) Straight line parallel to time axis
c) Exponential
d) Parabola
Answer<\/strong>: b
Explanation<\/strong>: Total energy remains constant, hence horizontal straight line.<\/p>\n\n\n\n

45. The time taken to move from equilibrium to extreme position is:
<\/mark><\/strong>a) T
b) T\/2
c) T\/4
d) T\/8
Answer:<\/strong> c
Explanation<\/strong>: Quarter of the time period is required to move from mean to extreme position.<\/p>\n\n\n\n

46. The displacement at which KE = PE in SHM is:
<\/mark><\/strong>a) A\/2
b) A\/\u221a2
c) A
d) 0
Answer:<\/strong> b
Explanation:<\/strong> At KE = PE, x=A\/\u221a2.<\/p>\n\n\n\n

47. The average kinetic energy in one time period of SHM is:
<\/mark><\/strong>a) Zero
b) E\/2
c) E
d) E\/4
Answer: b
Explanation: On average, KE = PE = half of total energy.<\/p>\n\n\n\n

48. The average potential energy in one time period of SHM is:
<\/mark><\/strong>a) Zero
b) E\/2
c) E
d) E\/4
Answer: b
Explanation: Same as KE, average PE = half of total energy.<\/p>\n\n\n\n

49. The average total energy in SHM over one cycle is:
<\/mark><\/strong>a) Zero
b) E
c) E\/2
d) 2E
Answer<\/strong>: b
Explanation<\/strong>: Total energy remains constant at E.<\/p>\n\n\n\n

50. In SHM, the velocity is zero for what fraction of the time period?
<\/mark><\/strong>a) Half
b) One-fourth
c) Twice in a cycle (at extremes)
d) Always
Answer<\/strong>: c
Explanation<\/strong>: Velocity = 0 at extreme positions, occurs twice in each cycle.

51. The projection of uniform circular motion on a diameter of the circle represents:
<\/mark><\/strong>a) Rectilinear motion
b) Simple harmonic motion
c) Projectile motion
d) Random motion
Answer<\/strong>: b
Explanation<\/strong>: SHM can be considered as the projection of uniform circular motion on one of its diameters.<\/p>\n\n\n\n

52. A body of mass mattached to a spring has time period T. If mass is doubled, the new time period will be:
<\/mark><\/strong>a) T\/2
b) T\u221a2
c) T\/\u221a2
d) 2T
Answer<\/strong>: b
Explanation<\/strong>: T=2\u03c0\u221a(m\/k). If mass doubles \u2192 T^’=2\u03c0\u221a(2m\/k)=T\u221a2.<\/p>\n\n\n\n

53. If spring constant kis doubled, the time period of SHM becomes:
<\/mark><\/strong>a) T\/2
b) T\/\u221a2
c) T\u221a2
d) 2T
Answer<\/strong>: b
Explanation<\/strong>: T=2\u03c0\u221a(m\/k). If k\u21922k, then T^’=T\/\u221a2.<\/p>\n\n\n\n

54. The length of a simple pendulum is increased 4 times. Its time period becomes:
<\/mark><\/strong>a) 2 times
b) 4 times
c) Half
d) Quarter
Answer<\/strong>: a
Explanation<\/strong>: T=2\u03c0\u221a(l\/g). If length increases 4\u00d7 \u2192 Tincreases 2\u00d7.<\/p>\n\n\n\n

55. A particle executes SHM with amplitude 10 cm. Its displacement when potential energy is 25% of total energy is:
<\/mark><\/strong>a) 5 cm
b) 10 cm
c) 10\/\u221a2cm
d) 10\/2\u221a2cm
Answer<\/strong>: a
Explanation<\/strong>: U\/E=x^2\/A^2=1\/4. So x=A\/2=5cm.<\/p>\n\n\n\n

56. A pendulum has a period of 2 s on Earth. On the Moon (g_moon=g\/6), the period will be:
<\/mark><\/strong>a) 2 s
b) 2\/\u221a6s
c) 2\u221a6s
d) 12s
Answer<\/strong>: c
Explanation<\/strong>: T\u221d1\/\u221ag. On moon, T^’=T\u221a(g\/g_moon )=2\u221a6.<\/p>\n\n\n\n

57. A 2 kg mass attached to a spring has amplitude 5 cm and frequency 2 Hz. Maximum velocity is:
<\/mark><\/strong>a) 0.2″\u2009m\/s”
b) 0.5″\u2009m\/s”
c) 0.628″\u2009m\/s”
d) 1.256″\u2009m\/s”
Answer<\/strong>: d
Explanation<\/strong>: v_max=\u03c9A. Here \u03c9=2\u03c0f=4\u03c0, A=0.05. So v=4\u03c0\u00d70.05=0.628m\/s. Wait, with f=2 Hz \u2192 \u03c9=4\u03c0, v_max=4\u03c0\u00d70.05\u22480.628. Correct answer = c.<\/p>\n\n\n\n

58. In SHM, the time period of oscillation of a spring mass system is:
<\/mark><\/strong>a) 2\u03c0\u221a(m\/k)
b) 2\u03c0\u221a(k\/m)
c) 2\u03c0\u221a(mg\/k)
d) 2\u03c0\u221a(g\/k)
Answer<\/strong>: a
Explanation<\/strong>: T=2\u03c0\u221a(m\/k).<\/p>\n\n\n\n

59. The maximum acceleration in SHM occurs at:
<\/mark><\/strong>a) Mean position
b) Extreme positions
c) Half amplitude
d) Everywhere
Answer<\/strong>: b
Explanation<\/strong>: a=-\u03c9^2 x. Maximum at maximum x.<\/p>\n\n\n\n

60. A 1 m long pendulum makes 60 oscillations per minute. The value of g is:
<\/mark><\/strong>a) 9.8 m\/s\u00b2
b) 9.9 m\/s\u00b2
c) 10.0 m\/s\u00b2
d) 10.2 m\/s\u00b2
Answer<\/strong>: a
Explanation<\/strong>: Frequency = 60\/60 = 1 Hz, T=1. T=2\u03c0\u221a(l\/g). Solve: 1=2\u03c0\u221a(1\/g)\u21d2g\u22489.8.<\/p>\n\n\n\n

61. A spring of force constant 200 N\/m is stretched by 0.1 m. The potential energy stored is:
<\/mark><\/strong>a) 1 J
b) 0.5 J
c) 2 J
d) 10 J
Answer<\/strong>: b
Explanation<\/strong>: U=1\/2 kx^2=0.5\u00d7200\u00d7\u30160.1\u3017^2=1. Correction: actually = 1 J. Correct answer = a.<\/p>\n\n\n\n

62. If frequency of SHM is doubled, the angular frequency becomes:
<\/mark><\/strong>a) Double
b) Half
c) Four times
d) Same
Answer<\/strong>: a
Explanation<\/strong>: \u03c9=2\u03c0f. If fdoubles, \u03c9doubles.<\/p>\n\n\n\n

63. A 2 kg mass is attached to a spring of constant k=50″\u2009N\/m” . Its time period is:
<\/mark><\/strong>a) 1.25 s
b) 2 s
c) 0.5 s
d) 4 s
Answer<\/strong>: a
Explanation<\/strong>: T=2\u03c0\u221a(m\/k)=2\u03c0\u221a(2\/50)=2\u03c0\u00d70.2=1.26\u22481.25.<\/p>\n\n\n\n

64. A particle executes SHM of amplitude 5 cm. At mean position, its velocity is 10 cm\/s. Time period is:
<\/mark><\/strong>a) \u03c0s
b) 2\u03c0s
c) \u03c0\/2s
d) 1 s
Answer<\/strong>: a
Explanation<\/strong>: v_max=\u03c9A. 10=\u03c9\u00d75. \u03c9=2, T=2\u03c0\/\u03c9=\u03c0.<\/p>\n\n\n\n

65. A mass-spring system oscillates with period 2 s. If mass is increased by 3 times, new period is:
<\/mark><\/strong>a) 2 s
b) 4 s
c) 6 s
d) 2\u221a3s
Answer<\/strong>: d
Explanation<\/strong>: T\u221d\u221am. New T=2\u221a3.<\/p>\n\n\n\n

66. If the displacement in SHM is halved, acceleration becomes:
<\/mark><\/strong>a) Half
b) Double
c) Same
d) One-fourth
Answer<\/strong>: a
Explanation<\/strong>: a=-\u03c9^2 x. If xhalves, ahalves.<\/p>\n\n\n\n

67. The energy distribution between KE and PE in SHM is:
<\/mark><\/strong>a) Constant
b) Alternating exchange
c) Always equal
d) Zero
Answer<\/strong>: b
Explanation<\/strong>: Energy oscillates between kinetic and potential forms.<\/p>\n\n\n\n

68. A pendulum clock keeps correct time at Earth. On the Moon it will:
<\/mark><\/strong>a) Run faster
b) Run slower
c) Run at same rate
d) Stop working
Answer<\/strong>: b
Explanation<\/strong>: T\u221d1\/\u221ag. On Moon, Tincreases \u2192 slower.<\/p>\n\n\n\n

69. The time period of a simple pendulum is proportional to:
<\/mark><\/strong>a) \u221al
b) 1\/\u221al
c) l
d) 1\/l
Answer<\/strong>: a
Explanation<\/strong>: T=2\u03c0\u221a(l\/g).<\/p>\n\n\n\n

70. For SHM, the restoring force is:
<\/mark><\/strong>a) Proportional to velocity
b) Proportional to displacement
c) Independent of displacement
d) Opposite to acceleration
Answer<\/strong>: b
Explanation<\/strong>: F=-kx.<\/p>\n\n\n\n

71. A pendulum of length 1 m has time period 2 s. If length is increased to 4 m, time period is:
<\/mark><\/strong>a) 2 s
b) 4 s
c) 6 s
d) 8 s
Answer<\/strong>: b
Explanation<\/strong>: T\u221d\u221al. So 2\u00d7 length factor \u2192 2\u00d7 period = 4 s.<\/p>\n\n\n\n

72. Which parameter remains constant in SHM?
<\/mark><\/strong>a) Amplitude
b) Velocity
c) Displacement
d) Acceleration
Answer<\/strong>: a
Explanation<\/mark><\/strong>: Amplitude is fixed (if no damping).<\/p>\n\n\n\n

73. A particle in SHM has amplitude 0.1 m and angular frequency 5 rad\/s. Maximum acceleration is:
<\/mark><\/strong>a) 0.5 m\/s\u00b2
b) 2.5 m\/s\u00b2
c) 5 m\/s\u00b2
d) 10 m\/s\u00b2
Answer<\/strong>: b
Explanation<\/strong>: a_max=\u03c9^2 A=25\u00d70.1=2.5.<\/p>\n\n\n\n

74. A spring has time period 2 s. If spring constant is made 4 times, new period is:
<\/mark><\/strong>a) 2 s
b) 1 s
c) 0.5 s
d) 4 s
Answer<\/strong>: b
Explanation<\/strong>: T\u221d1\/\u221ak. 4\u00d7 k \u2192 T\/2= 1 s.<\/p>\n\n\n\n

75. If the time period of SHM is 2 s, the time taken to move from mean position to extreme is:
<\/mark><\/strong>a) 0.5 s
b) 1 s
c) 2 s
d) 4 s
Answer<\/strong>: a
Explanation<\/strong>: Quarter of time period = 0.5 s.
76. The period of a simple pendulum is maximum at:
<\/mark><\/strong>a) Earth\u2019s poles
b) Earth\u2019s equator
c) On the Moon
d) At the center of Earth
Answer<\/strong>: d
Explanation<\/strong>: T=2\u03c0\u221a(l\/g). At Earth\u2019s center, g=0, so theoretically T\u2192\u221e.<\/p>\n\n\n\n

77. A body performing SHM passes through mean position. At that instant:
<\/mark><\/strong>a) Velocity is zero, acceleration is maximum
b) Velocity is maximum, acceleration is zero
c) Velocity and acceleration both maximum
d) Velocity and acceleration both zero
Answer<\/strong>: b
Explanation<\/strong>: At mean position, displacement = 0 \u2192 acceleration = 0, velocity = maximum.<\/p>\n\n\n\n

78. If total energy of SHM is 20 J and amplitude is doubled, new energy is:
<\/mark><\/strong>a) 20 J
b) 40 J
c) 80 J
d) 160 J
Answer<\/strong>: c
Explanation<\/strong>: Energy \u221d A^2. Doubling amplitude \u2192 4\u00d7 energy = 80 J.<\/p>\n\n\n\n

79. A body in SHM has amplitude 0.2 m and maximum velocity 0.4 m\/s. Its time period is:
<\/mark><\/strong>a) 1 s
b) 2 s
c) \u03c0s
d) 2\u03c0s
Answer<\/strong>: d
Explanation<\/strong>: v_max=\u03c9A. \u03c9=v_max\/A=0.4\/0.2=2. T=2\u03c0\/\u03c9=2\u03c0\/2=\u03c0. Correct answer = c.<\/p>\n\n\n\n

80. The motion of the second hand of a clock is:
<\/mark><\/strong>a) SHM
b) Uniform circular motion
c) Non-uniform circular motion
d) Linear motion
Answer<\/strong>: b
Explanation<\/strong>: It moves in uniform circular motion, not SHM.<\/p>\n\n\n\n

81. In SHM, the velocity is maximum at:<\/mark><\/strong>
a) Extreme positions
b) Midpoint (mean position)
c) Halfway displacement
d) Everywhere
Answer: b
Explanation: Velocity is maximum at mean position.<\/p>\n\n\n\n

82. The pendulum of a wall clock is made shorter. The clock will:
<\/mark><\/strong>a) Lose time
b) Gain time
c) Run at same rate
d) Stop
Answer<\/strong>: b
Explanation<\/strong>: T\u221d\u221al. Shorter length \u2192 smaller period \u2192 faster oscillations \u2192 clock gains time.<\/p>\n\n\n\n

83. A spring-mass oscillator has time period 2 s. If both mass and spring constant are doubled, the new period is:
<\/mark><\/strong>a) 1 s
b) 2 s
c) 4 s
d) \u221a2s
Answer<\/strong>: b
Explanation<\/strong>: T=2\u03c0\u221a(m\/k). If both mand kdouble, ratio m\/kremains same, so period unchanged.<\/p>\n\n\n\n

84. The displacement in SHM is x=0.1sin\u2061(100t). The frequency is:
<\/mark><\/strong>a) 100 Hz
b) 50 Hz
c) 100\/2\u03c0Hz
d) 50\/\u03c0Hz
Answer<\/strong>: c
Explanation<\/strong>: \u03c9=100. f=\u03c9\/2\u03c0=100\/2\u03c0\u224815.9Hz.<\/p>\n\n\n\n

85. A pendulum clock is taken to the top of a mountain. The clock will:
<\/mark><\/strong>a) Run faster
b) Run slower
c) Run at same rate
d) Stop
Answer<\/strong>: b
Explanation<\/strong>: At higher altitude, gdecreases \u2192 Tincreases \u2192 clock runs slower.<\/p>\n\n\n\n

86. Which factor does not affect the period of a pendulum?
<\/mark><\/strong>a) Mass of bob
b) Length of pendulum
c) Acceleration due to gravity
d) Amplitude (small)
Answer<\/strong>: a
Explanation<\/strong>: Time period is independent of bob\u2019s mass.<\/p>\n\n\n\n

87. The time period of SHM depends upon:
<\/mark><\/strong>a) Mass and restoring force constant
b) Only amplitude
c) Only velocity
d) Both amplitude and velocity
Answer<\/strong>: a
Explanation<\/strong>: T=2\u03c0\u221a(m\/k).<\/p>\n\n\n\n

88. In SHM, displacement is maximum when:
<\/mark><\/strong>a) Velocity is maximum
b) Acceleration is zero
c) Velocity is zero
d) Energy is maximum
Answer:<\/strong> c
Explanation:<\/strong> At extreme positions, displacement is maximum and velocity = 0.<\/p>\n\n\n\n

89. A simple pendulum has length land period T. If length is decreased by 9 times, new period is:
<\/mark><\/strong>a) T\/9
b) T\/3
c) T\/2
d) 3T
Answer<\/strong>: b
Explanation<\/strong>: T\u221d\u221al. If length reduces 9\u00d7, T^’=T\/\u221a9=T\/3.<\/p>\n\n\n\n

90. The time period of a mass-spring system on the Moon compared to Earth is:
<\/mark><\/strong>a) Same
b) Smaller
c) Larger
d) Infinite
Answer<\/strong>: a
Explanation<\/strong>: T=2\u03c0\u221a(m\/k). Independent of g.<\/p>\n\n\n\n

91. The motion of electrons in atoms can be approximately considered as:
<\/mark><\/strong>a) Linear motion
b) Uniform motion
c) SHM
d) None
Answer<\/strong>: c
Explanation<\/strong>: Small oscillations of electrons about equilibrium can be approximated as SHM.<\/p>\n\n\n\n

92. If energy in SHM is Eand amplitude is doubled, the new energy is:
<\/mark><\/strong>a) E
b) 2E
c) 3E
d) 4E
Answer<\/strong>: d
Explanation<\/strong>: E\u221dA^2. Doubling amplitude \u2192 4E.<\/p>\n\n\n\n

93. Which of the following oscillates with SHM?
<\/mark><\/strong>a) Tuning fork prongs
b) Pendulum bob (small oscillations)
c) Spring mass system
d) All of these
Answer<\/strong>: d
Explanation<\/strong>: All are SHM examples.<\/p>\n\n\n\n

94. The restoring force of SHM is provided by:
<\/mark><\/strong>a) Inertia
b) Elasticity or gravity
c) Friction
d) Centripetal force
Answer<\/strong>: b
Explanation<\/strong>: Restoring force comes from elasticity (spring) or gravity (pendulum).<\/p>\n\n\n\n

95. The displacement-time graph of SHM is:
<\/mark><\/strong>a) Straight line
b) Sine curve
c) Exponential
d) Hyperbola
Answer<\/strong>: b
Explanation<\/strong>: SHM displacement follows sinusoidal function.<\/p>\n\n\n\n

96. If a particle takes 0.5 s from mean to extreme, then its period is:
<\/mark><\/strong>a) 0.5 s
b) 1 s
c) 2 s
d) 4 s
Answer<\/strong>: c
Explanation<\/strong>: Mean to extreme = T\/4. So T=4\u00d70.5=2s.<\/p>\n\n\n\n

97. A body in SHM has amplitude 2 cm and frequency 50 Hz. Maximum acceleration is:
<\/strong><\/mark>a) 100″\u2009” \u3016”m\/s” \u3017^2
b) 200″\u2009” \u3016”m\/s” \u3017^2
c) 500″\u2009” \u3016”m\/s” \u3017^2
d) 2000″\u2009” \u3016”m\/s” \u3017^2
Answer<\/strong>: b
Explanation<\/strong>: a_max=\u03c9^2 A. \u03c9=2\u03c0f=314, A=0.02. a=(314^2)(0.02)\u22481970\u22482000. Correct = d.<\/p>\n\n\n\n

98. The potential energy in SHM varies with:
<\/mark><\/strong>a) Time linearly
b) Time sinusoidally
c) Displacement squared
d) Velocity
Answer<\/strong>: c
Explanation<\/strong>: U=1\/2 kx^2.<\/p>\n\n\n\n

99. The average value of displacement in SHM over one complete cycle is:
<\/mark><\/strong>a) Amplitude
b) Zero
c) Half amplitude
d) Infinite
Answer<\/strong>: b
Explanation<\/strong>: Displacement is symmetric, positive half cancels negative half.<\/p>\n\n\n\n

100. The average kinetic energy of SHM over one complete cycle is:
<\/mark><\/strong>a) Zero
b) Equal to total energy
c) Half of total energy
d) Twice total energy
Answer<\/strong>: c
Explanation<\/strong>: On average, KE = PE = half of total energy.<\/p>\n","protected":false},"excerpt":{"rendered":"

1. Simple harmonic motion is defined as:a) Motion with constant velocityb) Motion with acceleration proportional to displacement and directed opposite to itc) Motion with constant accelerationd) Motion with displacement proportional to velocityAnswer: bExplanation: SHM is a periodic motion where restoring force (or acceleration) is directly proportional to displacement and acts in the opposite direction, i.e.,<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[15668,15481,15671,15677,15669,15667,4029,5649,5652,15465,5623,15662,15673,15665,15676,15666,15674,15536,15479,15474,15467,15483,15472,15592,15456,15478,15514,15663,15675,15664,10954,15672,15670,15469,15480],"class_list":{"0":"post-12641","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-physics","7":"tag-amplitude","8":"tag-competitive-exam-physics","9":"tag-damped-oscillations","10":"tag-energy-in-shm","11":"tag-forced-oscillations","12":"tag-frequency-and-time-period","13":"tag-mcqs-adda","14":"tag-mcqs-for-pc-psi-sda-fda-pdo-vao-banking-kas-ias-ssc-gd-ssc-chsl-ssc-cgl-for-all-compitative-exams","15":"tag-mcqs-for-pc-psi-sda-fda-pdo-vao-banking-kas-ias-ssc-gd-ssc-chsl-ssc-cgl-for-all-compitative-examsin-kannada","16":"tag-mcqs-for-physics-exam","17":"tag-mcqs-for-sda-fda-pdo-vao-banking-kas-ias-ssc-gd-ssc-chsl-ssc-cgl-for-all-compitative-exams","18":"tag-oscillations","19":"tag-oscillatory-motion","20":"tag-pendulum-motion","21":"tag-period-of-oscillation","22":"tag-periodic-motion","23":"tag-phase-in-shm","24":"tag-physics-formulas","25":"tag-physics-learning","26":"tag-physics-mcqs","27":"tag-physics-preparation-material","28":"tag-physics-questions-and-answers","29":"tag-physics-quiz","30":"tag-physics-revision","31":"tag-physics-study-material","32":"tag-psc-physics-mcqs","33":"tag-resonance","34":"tag-shm-in-physics","35":"tag-shm-problems","36":"tag-simple-harmonic-motion","37":"tag-simple-harmonic-motion-top-100-mcqs-with-answer-and-explanation","38":"tag-simple-harmonic-oscillator","39":"tag-spring-mass-system","40":"tag-ssc-physics-mcqs","41":"tag-upsc-physics-mcqs"},"_links":{"self":[{"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/posts\/12641","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/comments?post=12641"}],"version-history":[{"count":2,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/posts\/12641\/revisions"}],"predecessor-version":[{"id":12665,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/posts\/12641\/revisions\/12665"}],"wp:attachment":[{"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/media?parent=12641"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/categories?post=12641"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/tags?post=12641"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}