1. Simple harmonic motion is defined as:
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: b
Explanation: SHM is a periodic motion where restoring force (or acceleration) is directly proportional to displacement and acts in the opposite direction, i.e., a∝-x.
2. The displacement in SHM is given by:
a) x=Asin(ωt+ϕ)
b) x=Acos(ωt+ϕ)
c) Both (a) and (b)
d) None of these
Answer: c
Explanation: Displacement in SHM can be represented by sine or cosine functions depending on initial conditions.
3. The time period of a simple harmonic oscillator depends on:
a) Amplitude only
b) Mass and restoring constant
c) Velocity of oscillation
d) None of these
Answer: b
Explanation: Time period T=2π√(m/k). It is independent of amplitude.
4. In SHM, acceleration is maximum when:
a) Displacement is zero
b) Displacement is maximum
c) Velocity is maximum
d) At equilibrium position
Answer: b
Explanation: a=-ω^2 x. Maximum acceleration occurs at maximum displacement.
5. Velocity in SHM is maximum when:
a) Displacement is zero
b) Displacement is maximum
c) Acceleration is maximum
d) At turning points
Answer: a
Explanation: At equilibrium (x=0), velocity is maximum.
6. The phase difference between velocity and displacement in SHM is:
a) 0^∘
b) 90^∘
c) 180^∘
d) 270^∘
Answer: b
Explanation: Velocity leads displacement by π/2in SHM.
7. Which of the following is an example of SHM?
a) Uniform circular motion projection
b) A pendulum oscillating with small amplitude
c) A mass attached to a spring
d) All of these
Answer: d
Explanation: All these are classic examples of SHM.
8. In SHM, kinetic energy is maximum when:
a) Displacement is maximum
b) Displacement is zero
c) Acceleration is maximum
d) None
Answer: b
Explanation: KE = maximum at equilibrium (displacement = 0).
9. Which of the following is true about potential energy in SHM?
a) Maximum at equilibrium
b) Minimum at extreme position
c) Maximum at extreme position
d) Constant everywhere
Answer: c
Explanation: Potential energy is maximum at maximum displacement.
10. A pendulum performs SHM only if:
a) Angle of oscillation is very small
b) Mass of bob is large
c) Length of string is small
d) For all cases
Answer: a
Explanation: For small angles, sinθ≈θ, making the motion approximately SHM.
11. The restoring force in SHM is proportional to:
a) Velocity
b) Displacement
c) Acceleration
d) Momentum
Answer: b
Explanation: Restoring force F=-kx.
12. If a body in SHM has amplitude A, the maximum velocity is:
a) ωA
b) A/ω
c) Aω^2
d) None
Answer: a
Explanation: v_”max” =ωA.
13. The total energy in SHM is proportional to:
a) Amplitude
b) Amplitude squared
c) Velocity
d) Frequency
Answer: b
Explanation: E=1/2 kA^2, proportional to A^2.
14. The frequency of SHM is:
a) Independent of amplitude
b) Directly proportional to amplitude
c) Inversely proportional to amplitude
d) Depends on energy
Answer: a
Explanation: Frequency depends on system parameters, not amplitude.
15. The graph of displacement vs time in SHM is:
a) Straight line
b) Exponential
c) Sinusoidal
d) Parabola
Answer: c
Explanation: Displacement is a sine or cosine function of time.
16. In SHM, velocity is zero when:
a) Acceleration is maximum
b) Displacement is zero
c) Energy is maximum
d) At equilibrium
Answer: a
Explanation: At extreme positions, acceleration is maximum, velocity is zero.
17. The time period of a simple pendulum is independent of:
a) Length of pendulum
b) Value of acceleration due to gravity
c) Mass of bob
d) Amplitude (small)
Answer: c
Explanation: T=2π√(l/g), mass has no effect.
18. The dimension of angular frequency ωis:
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: a
Explanation: Angular frequency has units of rad/s, i.e. T^(-1).
19. Which one is not an SHM?
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: c
Explanation: Earth’s motion is elliptical, not SHM.
20. The frequency of oscillation is reciprocal of:
a) Amplitude
b) Time period
c) Phase angle
d) Velocity
Answer: b
Explanation: Frequency = 1/T.
21. If the time period is doubled, the frequency becomes:
a) Double
b) Half
c) Same
d) Four times
Answer: b
Explanation: Frequency = inverse of time period.
22. In SHM, potential energy is zero when:
a) Displacement is maximum
b) Displacement is zero
c) Velocity is zero
d) Acceleration is maximum
Answer: b
Explanation: At equilibrium, potential energy is zero.
23. Which physical quantity remains constant in SHM?
a) Velocity
b) Acceleration
c) Total energy
d) Potential energy
Answer: c
Explanation: Total energy remains constant (conservation of energy).
24. SHM is a type of:
a) Uniform motion
b) Periodic motion
c) Rectilinear motion
d) Projectile motion
Answer: b
Explanation: SHM is a periodic oscillatory motion.
25. The graph between restoring force and displacement in SHM is:
a) Straight line passing through origin
b) Parabola
c) Exponential
d) Constant
Answer: a
Explanation: F=-kx, a straight line graph with negative slope.
26. The general equation of displacement in SHM is:
a) x=Asin(ωt)
b) x=Acos(ωt)
c) x=Asin(ωt+ϕ)
d) All of these
Answer: d
Explanation: Depending on initial conditions, SHM displacement can be expressed using sine, cosine, or with phase angle.
27. If displacement is x=Asin(ωt), then velocity is:
a) v=Aωcos(ωt)
b) v=-Aωsin(ωt)
c) v=Acos(ωt)
d) v=Asin(ωt)
Answer: a
Explanation: Velocity is derivative of displacement: v=dx/dt=Aωcos(ωt).
28. The maximum acceleration in SHM is:
a) Aω
b) Aω^2
c) ω^2/A
d) 1/(Aω^2)
Answer: b
Explanation: a=-ω^2 x. Maximum acceleration = ω^2 A.
29. The phase difference between acceleration and displacement in SHM is:
a) 0^∘
b) 90^∘
c) 180^∘
d) 270^∘
Answer: c
Explanation: Since a=-ω^2 x, acceleration is always opposite to displacement (phase difference π).
30. The kinetic energy of a particle in SHM is given by:
a) 1/2 kx^2
b) 1/2 mω^2 (A^2-x^2)
c) 1/2 mv^2
d) Both b and c
Answer: d
Explanation: KE in SHM can be expressed in terms of displacement or velocity.
31. The potential energy of SHM is:
a) 1/2 mω^2 x^2
b) 1/2 mv^2
c) 1/2 kv^2
d) 1/2 A^2
Answer: a
Explanation: Potential energy = work done by restoring force = 1/2 kx^2=1/2 mω^2 x^2.
32. The total energy in SHM is given by:
a) 1/2 mω^2 A^2
b) 1/2 mω^2 x^2
c) 1/2 mv^2
d) None
Answer: a
Explanation: Total mechanical energy = constant = 1/2 mω^2 A^2.
33. At equilibrium, in SHM:
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: a
Explanation: At equilibrium x=0, PE=0, KE = maximum.
34. At extreme positions, in SHM:
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: b
Explanation: At extreme displacement, velocity = 0, so KE=0, PE=maximum.
35. The ratio of kinetic energy to total energy in SHM is:
a) x^2/A^2
b) 1-x^2/A^2
c) A^2/x^2
d) Constant
Answer: b
Explanation: KE = 1/2 mω^2 (A^2-x^2), total = 1/2 mω^2 A^2. Ratio = 1-(x^2/A^2).
36. The ratio of potential energy to total energy in SHM is:
a) x^2/A^2
b) 1-x^2/A^2
c) Constant
d) Zero
Answer: a
Explanation: PE = 1/2 mω^2 x^2, total = 1/2 mω^2 A^2. Ratio = x^2/A^2.
37. The frequency of SHM depends on:
a) Amplitude
b) Mass and spring constant
c) Velocity
d) Energy
Answer: b
Explanation: For spring-mass system, f=1/2π √(k/m).
38. For a pendulum of length l, the time period is:
a) 2π√(g/l)
b) 2π√(l/g)
c) π√(l/g)
d) 2π l/g
Answer: b
Explanation: Time period of simple pendulum = T=2π√(l/g).
39. In SHM, the restoring force is always directed:
a) Along velocity
b) Opposite to velocity
c) Towards mean position
d) Away from mean position
Answer: c
Explanation: Restoring force always tries to bring particle back to equilibrium.
40. If the amplitude is doubled, the maximum velocity becomes:
a) Half
b) Double
c) Same
d) Four times
Answer: b
Explanation: v_max=ωA, so directly proportional to amplitude.
41. If amplitude is doubled, the total energy becomes:
a) Double
b) Same
c) Four times
d) Half
Answer: c
Explanation: Total energy ∝ A^2. So doubling amplitude → 4× energy.
42. At half the amplitude, the potential energy is:
a) 1/4of total energy
b) 1/2of total energy
c) 3/4of total energy
d) Equal to total energy
Answer: a
Explanation: U/E=(x^2/A^2)=(A/2)^2/A^2=1/4.
43. At half the amplitude, the kinetic energy is:
a) 1/4of total energy
b) 3/4of total energy
c) Equal to potential energy
d) Zero
Answer: b
Explanation: KE/E = 1 – (x^2/A^2) = 1 – 1/4 = 3/4.
44. The graph of total energy vs time in SHM is:
a) Sinusoidal
b) Straight line parallel to time axis
c) Exponential
d) Parabola
Answer: b
Explanation: Total energy remains constant, hence horizontal straight line.
45. The time taken to move from equilibrium to extreme position is:
a) T
b) T/2
c) T/4
d) T/8
Answer: c
Explanation: Quarter of the time period is required to move from mean to extreme position.
46. The displacement at which KE = PE in SHM is:
a) A/2
b) A/√2
c) A
d) 0
Answer: b
Explanation: At KE = PE, x=A/√2.
47. The average kinetic energy in one time period of SHM is:
a) Zero
b) E/2
c) E
d) E/4
Answer: b
Explanation: On average, KE = PE = half of total energy.
48. The average potential energy in one time period of SHM is:
a) Zero
b) E/2
c) E
d) E/4
Answer: b
Explanation: Same as KE, average PE = half of total energy.
49. The average total energy in SHM over one cycle is:
a) Zero
b) E
c) E/2
d) 2E
Answer: b
Explanation: Total energy remains constant at E.
50. In SHM, the velocity is zero for what fraction of the time period?
a) Half
b) One-fourth
c) Twice in a cycle (at extremes)
d) Always
Answer: c
Explanation: Velocity = 0 at extreme positions, occurs twice in each cycle.
51. The projection of uniform circular motion on a diameter of the circle represents:
a) Rectilinear motion
b) Simple harmonic motion
c) Projectile motion
d) Random motion
Answer: b
Explanation: SHM can be considered as the projection of uniform circular motion on one of its diameters.
52. A body of mass mattached to a spring has time period T. If mass is doubled, the new time period will be:
a) T/2
b) T√2
c) T/√2
d) 2T
Answer: b
Explanation: T=2π√(m/k). If mass doubles → T^’=2π√(2m/k)=T√2.
53. If spring constant kis doubled, the time period of SHM becomes:
a) T/2
b) T/√2
c) T√2
d) 2T
Answer: b
Explanation: T=2π√(m/k). If k→2k, then T^’=T/√2.
54. The length of a simple pendulum is increased 4 times. Its time period becomes:
a) 2 times
b) 4 times
c) Half
d) Quarter
Answer: a
Explanation: T=2π√(l/g). If length increases 4× → Tincreases 2×.
55. A particle executes SHM with amplitude 10 cm. Its displacement when potential energy is 25% of total energy is:
a) 5 cm
b) 10 cm
c) 10/√2cm
d) 10/2√2cm
Answer: a
Explanation: U/E=x^2/A^2=1/4. So x=A/2=5cm.
56. A pendulum has a period of 2 s on Earth. On the Moon (g_moon=g/6), the period will be:
a) 2 s
b) 2/√6s
c) 2√6s
d) 12s
Answer: c
Explanation: T∝1/√g. On moon, T^’=T√(g/g_moon )=2√6.
57. A 2 kg mass attached to a spring has amplitude 5 cm and frequency 2 Hz. Maximum velocity is:
a) 0.2″ m/s”
b) 0.5″ m/s”
c) 0.628″ m/s”
d) 1.256″ m/s”
Answer: d
Explanation: v_max=ωA. Here ω=2πf=4π, A=0.05. So v=4π×0.05=0.628m/s. Wait, with f=2 Hz → ω=4π, v_max=4π×0.05≈0.628. Correct answer = c.
58. In SHM, the time period of oscillation of a spring mass system is:
a) 2π√(m/k)
b) 2π√(k/m)
c) 2π√(mg/k)
d) 2π√(g/k)
Answer: a
Explanation: T=2π√(m/k).
59. The maximum acceleration in SHM occurs at:
a) Mean position
b) Extreme positions
c) Half amplitude
d) Everywhere
Answer: b
Explanation: a=-ω^2 x. Maximum at maximum x.
60. A 1 m long pendulum makes 60 oscillations per minute. The value of g is:
a) 9.8 m/s²
b) 9.9 m/s²
c) 10.0 m/s²
d) 10.2 m/s²
Answer: a
Explanation: Frequency = 60/60 = 1 Hz, T=1. T=2π√(l/g). Solve: 1=2π√(1/g)⇒g≈9.8.
61. A spring of force constant 200 N/m is stretched by 0.1 m. The potential energy stored is:
a) 1 J
b) 0.5 J
c) 2 J
d) 10 J
Answer: b
Explanation: U=1/2 kx^2=0.5×200×〖0.1〗^2=1. Correction: actually = 1 J. Correct answer = a.
62. If frequency of SHM is doubled, the angular frequency becomes:
a) Double
b) Half
c) Four times
d) Same
Answer: a
Explanation: ω=2πf. If fdoubles, ωdoubles.
63. A 2 kg mass is attached to a spring of constant k=50″ N/m” . Its time period is:
a) 1.25 s
b) 2 s
c) 0.5 s
d) 4 s
Answer: a
Explanation: T=2π√(m/k)=2π√(2/50)=2π×0.2=1.26≈1.25.
64. A particle executes SHM of amplitude 5 cm. At mean position, its velocity is 10 cm/s. Time period is:
a) πs
b) 2πs
c) π/2s
d) 1 s
Answer: a
Explanation: v_max=ωA. 10=ω×5. ω=2, T=2π/ω=π.
65. A mass-spring system oscillates with period 2 s. If mass is increased by 3 times, new period is:
a) 2 s
b) 4 s
c) 6 s
d) 2√3s
Answer: d
Explanation: T∝√m. New T=2√3.
66. If the displacement in SHM is halved, acceleration becomes:
a) Half
b) Double
c) Same
d) One-fourth
Answer: a
Explanation: a=-ω^2 x. If xhalves, ahalves.
67. The energy distribution between KE and PE in SHM is:
a) Constant
b) Alternating exchange
c) Always equal
d) Zero
Answer: b
Explanation: Energy oscillates between kinetic and potential forms.
68. A pendulum clock keeps correct time at Earth. On the Moon it will:
a) Run faster
b) Run slower
c) Run at same rate
d) Stop working
Answer: b
Explanation: T∝1/√g. On Moon, Tincreases → slower.
69. The time period of a simple pendulum is proportional to:
a) √l
b) 1/√l
c) l
d) 1/l
Answer: a
Explanation: T=2π√(l/g).
70. For SHM, the restoring force is:
a) Proportional to velocity
b) Proportional to displacement
c) Independent of displacement
d) Opposite to acceleration
Answer: b
Explanation: F=-kx.
71. A pendulum of length 1 m has time period 2 s. If length is increased to 4 m, time period is:
a) 2 s
b) 4 s
c) 6 s
d) 8 s
Answer: b
Explanation: T∝√l. So 2× length factor → 2× period = 4 s.
72. Which parameter remains constant in SHM?
a) Amplitude
b) Velocity
c) Displacement
d) Acceleration
Answer: a
Explanation: Amplitude is fixed (if no damping).
73. A particle in SHM has amplitude 0.1 m and angular frequency 5 rad/s. Maximum acceleration is:
a) 0.5 m/s²
b) 2.5 m/s²
c) 5 m/s²
d) 10 m/s²
Answer: b
Explanation: a_max=ω^2 A=25×0.1=2.5.
74. A spring has time period 2 s. If spring constant is made 4 times, new period is:
a) 2 s
b) 1 s
c) 0.5 s
d) 4 s
Answer: b
Explanation: T∝1/√k. 4× k → T/2= 1 s.
75. If the time period of SHM is 2 s, the time taken to move from mean position to extreme is:
a) 0.5 s
b) 1 s
c) 2 s
d) 4 s
Answer: a
Explanation: Quarter of time period = 0.5 s.
76. The period of a simple pendulum is maximum at:
a) Earth’s poles
b) Earth’s equator
c) On the Moon
d) At the center of Earth
Answer: d
Explanation: T=2π√(l/g). At Earth’s center, g=0, so theoretically T→∞.
77. A body performing SHM passes through mean position. At that instant:
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: b
Explanation: At mean position, displacement = 0 → acceleration = 0, velocity = maximum.
78. If total energy of SHM is 20 J and amplitude is doubled, new energy is:
a) 20 J
b) 40 J
c) 80 J
d) 160 J
Answer: c
Explanation: Energy ∝ A^2. Doubling amplitude → 4× energy = 80 J.
79. A body in SHM has amplitude 0.2 m and maximum velocity 0.4 m/s. Its time period is:
a) 1 s
b) 2 s
c) πs
d) 2πs
Answer: d
Explanation: v_max=ωA. ω=v_max/A=0.4/0.2=2. T=2π/ω=2π/2=π. Correct answer = c.
80. The motion of the second hand of a clock is:
a) SHM
b) Uniform circular motion
c) Non-uniform circular motion
d) Linear motion
Answer: b
Explanation: It moves in uniform circular motion, not SHM.
81. In SHM, the velocity is maximum at:
a) Extreme positions
b) Midpoint (mean position)
c) Halfway displacement
d) Everywhere
Answer: b
Explanation: Velocity is maximum at mean position.
82. The pendulum of a wall clock is made shorter. The clock will:
a) Lose time
b) Gain time
c) Run at same rate
d) Stop
Answer: b
Explanation: T∝√l. Shorter length → smaller period → faster oscillations → clock gains time.
83. A spring-mass oscillator has time period 2 s. If both mass and spring constant are doubled, the new period is:
a) 1 s
b) 2 s
c) 4 s
d) √2s
Answer: b
Explanation: T=2π√(m/k). If both mand kdouble, ratio m/kremains same, so period unchanged.
84. The displacement in SHM is x=0.1sin(100t). The frequency is:
a) 100 Hz
b) 50 Hz
c) 100/2πHz
d) 50/πHz
Answer: c
Explanation: ω=100. f=ω/2π=100/2π≈15.9Hz.
85. A pendulum clock is taken to the top of a mountain. The clock will:
a) Run faster
b) Run slower
c) Run at same rate
d) Stop
Answer: b
Explanation: At higher altitude, gdecreases → Tincreases → clock runs slower.
86. Which factor does not affect the period of a pendulum?
a) Mass of bob
b) Length of pendulum
c) Acceleration due to gravity
d) Amplitude (small)
Answer: a
Explanation: Time period is independent of bob’s mass.
87. The time period of SHM depends upon:
a) Mass and restoring force constant
b) Only amplitude
c) Only velocity
d) Both amplitude and velocity
Answer: a
Explanation: T=2π√(m/k).
88. In SHM, displacement is maximum when:
a) Velocity is maximum
b) Acceleration is zero
c) Velocity is zero
d) Energy is maximum
Answer: c
Explanation: At extreme positions, displacement is maximum and velocity = 0.
89. A simple pendulum has length land period T. If length is decreased by 9 times, new period is:
a) T/9
b) T/3
c) T/2
d) 3T
Answer: b
Explanation: T∝√l. If length reduces 9×, T^’=T/√9=T/3.
90. The time period of a mass-spring system on the Moon compared to Earth is:
a) Same
b) Smaller
c) Larger
d) Infinite
Answer: a
Explanation: T=2π√(m/k). Independent of g.
91. The motion of electrons in atoms can be approximately considered as:
a) Linear motion
b) Uniform motion
c) SHM
d) None
Answer: c
Explanation: Small oscillations of electrons about equilibrium can be approximated as SHM.
92. If energy in SHM is Eand amplitude is doubled, the new energy is:
a) E
b) 2E
c) 3E
d) 4E
Answer: d
Explanation: E∝A^2. Doubling amplitude → 4E.
93. Which of the following oscillates with SHM?
a) Tuning fork prongs
b) Pendulum bob (small oscillations)
c) Spring mass system
d) All of these
Answer: d
Explanation: All are SHM examples.
94. The restoring force of SHM is provided by:
a) Inertia
b) Elasticity or gravity
c) Friction
d) Centripetal force
Answer: b
Explanation: Restoring force comes from elasticity (spring) or gravity (pendulum).
95. The displacement-time graph of SHM is:
a) Straight line
b) Sine curve
c) Exponential
d) Hyperbola
Answer: b
Explanation: SHM displacement follows sinusoidal function.
96. If a particle takes 0.5 s from mean to extreme, then its period is:
a) 0.5 s
b) 1 s
c) 2 s
d) 4 s
Answer: c
Explanation: Mean to extreme = T/4. So T=4×0.5=2s.
97. A body in SHM has amplitude 2 cm and frequency 50 Hz. Maximum acceleration is:
a) 100″ ” 〖”m/s” 〗^2
b) 200″ ” 〖”m/s” 〗^2
c) 500″ ” 〖”m/s” 〗^2
d) 2000″ ” 〖”m/s” 〗^2
Answer: b
Explanation: a_max=ω^2 A. ω=2πf=314, A=0.02. a=(314^2)(0.02)≈1970≈2000. Correct = d.
98. The potential energy in SHM varies with:
a) Time linearly
b) Time sinusoidally
c) Displacement squared
d) Velocity
Answer: c
Explanation: U=1/2 kx^2.
99. The average value of displacement in SHM over one complete cycle is:
a) Amplitude
b) Zero
c) Half amplitude
d) Infinite
Answer: b
Explanation: Displacement is symmetric, positive half cancels negative half.
100. The average kinetic energy of SHM over one complete cycle is:
a) Zero
b) Equal to total energy
c) Half of total energy
d) Twice total energy
Answer: c
Explanation: On average, KE = PE = half of total energy.