100 High-Scoring MCQs on Oscillatory Motion (Set - 1 Basic Level MCQs)

100 High-Scoring MCQs on Oscillatory Motion (Set - 1 Basic Level MCQs)

100 High-Scoring MCQs (Set-1 Basic Level) on Chapter: Oscillation, Physics (Unit-Wise Practice):


This post contains 100 carefully selected multiple-choice questions (MCQs) (Set-1 Basic Level), on Oscillations, designed strictly according to all board exam syllabus. These MCQs include a balanced mix of conceptual and numerical problems, making them exam-ready, revision-friendly, and high-scoring.

Whether you are preparing for annual board examinations, chapter tests, or competitive entry tests, this comprehensive MCQ collection thoroughly covers all the fundamental concepts of oscillatory motion, simple harmonic motion, resonance, and damping. It helps students build a strong conceptual base and improve numerical problem-solving skills with confidence.
📘 This unit-wise MCQ set includes questions from:
  • Simple examples of free oscillations
  • Necessary conditions for simple harmonic motion (SHM)
  • SHM as the projection of uniform circular motion
  • Definitions of amplitude, period, frequency, angular frequency, and phase difference
  • Relationship of period with frequency and angular frequency
  • Defining equation of SHM a = - ⍵² x
  • Proof that mass attached to a spring performs SHM
  • Interchange of kinetic and potential energy during SHM
  • Motion of simple pendulum as SHM and its time period
  • Practical examples of free and forced oscillations (resonance)
  • Frequency response curve and amplitude variation near natural frequency
  • Practical examples of damped oscillations
  • Degree of damping and critical damping in car suspension systems
  • Factors affecting frequency response and sharpness of resonance
👉 Each MCQ is provided with:
  • Correct answer
  • Clear and concise explanation
👉 This MCQ set helps students to:
  • Strengthen conceptual understanding of oscillations and SHM
  • Master formulas and numerical problem-solving techniques
  • Understand resonance, damping, and energy exchange in oscillatory systems
  • Avoid common exam mistakes in conceptual and numerical questions
  • Score maximum marks in the MCQs section of board exams

This all-in-one MCQ collection is an essential resource for all students who want to fully master oscillatory motion concepts and excel in physics examinations.

MCQs No. 1

1. A motion that repeats itself after equal intervals of time is called:

a) Uniform motion
b) Periodic motion
c) Random motion
d) Linear motion

Correct Answer is option b. Periodic motion

Explanation: A periodic motion repeats itself after equal intervals of time.

2. Free oscillations occur when:

a) An external force acts continuously
b) No external force acts after initial disturbance
c) Friction is very large
d) The object moves in a circle

Correct Answer is option b. No external force acts after initial disturbance

Explanation: Free oscillations occur when the system oscillates naturally without any driving force.

3. The necessary condition for simple harmonic motion is that:

a) Velocity must be constant
b) Acceleration must be constant
c) Restoring force must be proportional to displacement
d) Displacement must be constant

Correct Answer is option c. Restoring force must be proportional to displacement

Explanation: SHM occurs only when the restoring force is directly proportional to displacement and opposite in direction.

4. The projection of uniform circular motion on a diameter is:

a) Uniform motion
b) Random motion
c) Simple harmonic motion
d) Circular motion

Correct Answer is option c. Simple harmonic motion

Explanation: Projection of uniform circular motion on a diameter gives SHM.

5. Amplitude of oscillation is defined as:

a) Time for one oscillation
b) Maximum displacement from mean position
c) Number of oscillations per second
d) Total distance travelled

Correct Answer is option b. Maximum displacement from mean position

Explanation: Amplitude is the maximum distance from equilibrium position.

6. The time taken to complete one oscillation is called:

a) Frequency
b) Period
c) Angular velocity
d) Phase

Correct Answer is option b. Period

Explanation: Period is the time for one complete vibration.

7. Frequency is related to period by:

a) f = T
b) f = 1/T
c) f = T²
d) f = 2T

Correct Answer is option b. f = 1/T

Explanation: Frequency is the reciprocal of period.

8. Angular frequency is given by:

a) ω = f/T
b) ω = 2πf
c) ω = T/f
d) ω = f²

Correct Answer is option b. ω = 2πf

Explanation: Angular frequency is related to frequency by ω = 2πf.

9. The defining equation of SHM is:

a) v = ωx
b) a = ωx
c) a = −ω²x
d) F = ma

Correct Answer is option c. a = −ω²x

Explanation: In SHM acceleration is proportional to displacement and opposite in direction.

10. The negative sign in a = −ω²x indicates that acceleration is:

a) Along displacement
b) Zero
c) Opposite to displacement
d) Constant

Correct Answer is option c. Opposite to displacement

Explanation: Restoring acceleration always acts toward mean position.

11. A mass attached to a spring executes SHM because:

a) Spring has no mass
b) Spring obeys Hooke’s law
c) Gravity is zero
d) Velocity is constant

Correct Answer is option b. Spring obeys Hooke’s law

Explanation: Hooke’s law ensures restoring force proportional to displacement.

12. Hooke’s law states that:

a) Force ∝ velocity
b) Force ∝ acceleration
c) Force ∝ displacement
d) Force ∝ time

Correct Answer is option c. Force ∝ displacement

Explanation: According to Hooke’s law, F = −kx.

13. In SHM, kinetic energy is maximum at:

a) Extreme position
b) Mean position
c) Midway
d) Everywhere same

Correct Answer is option b. Mean position

Explanation: Velocity is maximum at mean position, so KE is maximum.

14. Potential energy in SHM is maximum at:

a) Mean position
b) Extreme position
c) Everywhere zero
d) Midway

Correct Answer is option b. Extreme position

Explanation: Displacement is maximum at extremes, so PE is maximum.

15. In SHM, total energy is:

a) Increasing
b) Decreasing
c) Constant
d) Zero

Correct Answer is option c. Constant

Explanation: In ideal SHM, energy is conserved.

16. A simple pendulum performs SHM when:

a) Amplitude is large
b) Length is very small
c) Angular displacement is small
d) Gravity is zero

Correct Answer is option c. Angular displacement is small

Explanation: Pendulum behaves as SHM only for small angles.

17. Time period of a simple pendulum is:

a) T = 2π√(m/k)
b) T = 2π√(L/g)
c) T = 1/f
d) T = ω²

Correct Answer is option b. T = 2π√(L/g)

Explanation: This is the standard formula for pendulum period.

18. Forced oscillations occur when:

a) No external force acts
b) Object oscillates naturally
c) External periodic force acts
d) Friction is zero

Correct Answer is option c. External periodic force acts

Explanation: Forced oscillations are produced by an external driving force.

19. Resonance occurs when:

a) Driving frequency is zero
b) Natural frequency equals driving frequency
c) Amplitude is zero
d) Damping is maximum

Correct Answer is option b. Natural frequency equals driving frequency

Explanation: Resonance happens when driving frequency matches natural frequency.

20. Damped oscillations occur due to:

a) Gravity
b) External force
c) Friction or resistance
d) Spring constant

Correct Answer is option c. Friction or resistance

Explanation: Damping is caused by resistive forces like friction or air resistance.

21. The SI unit of frequency is:

a) Second
b) Hertz
c) Radian
d) Meter

Correct Answer is option b. Hertz

Explanation: Frequency is measured in hertz (Hz).

22. The SI unit of angular frequency is:

a) Hz
b) s⁻¹
c) rad s⁻¹
d) m s⁻¹

Correct Answer is option c. rad s⁻¹

Explanation: Angular frequency is measured in radians per second.

23. If frequency of oscillation is 5 Hz, its period is:

a) 5 s
b) 0.5 s
c) 0.2 s
d) 25 s

Correct Answer is option c. 0.2 s

Explanation: T = 1/f = 1/5 = 0.2 s.

24. If angular frequency is 4π rad s⁻¹, the frequency is:

a) 1 Hz
b) 2 Hz
c) 4 Hz
d) 8 Hz

Correct Answer is option b. 2 Hz

Explanation: f = ω / 2π = 4π / 2π = 2 Hz.

25. Phase difference is the difference in:

a) Speed
b) Displacement
c) Time period
d) Phase angle

Correct Answer is option d. Phase angle

Explanation: Phase difference is the difference in phase angles of two oscillations.

26. Two oscillations are said to be in phase if their phase difference is:

a) 90°
b) 180°
c) 0°
d) 270°

Correct Answer is option c. 0°

Explanation: In-phase oscillations have zero phase difference.

27. For a spring-mass system, if spring constant increases, time period:

a) Increases
b) Decreases
c) Remains same
d) Becomes zero

Correct Answer is option b. Decreases

Explanation: T = 2π√(m/k), so T decreases when k increases.

28. If the mass attached to a spring is doubled, the period becomes:

a) T/2
b) 2T
c) √2T
d) T/√2

Correct Answer is option c. √2T

Explanation: T ∝ √m, so doubling m makes T = √2T.

29. In SHM, velocity is maximum at:

a) Extreme position
b) Mean position
c) Everywhere same
d) Midway

Correct Answer is option b. Mean position

Explanation: Velocity is maximum at equilibrium position.

30. In SHM, acceleration is maximum at:

a) Mean position
b) Extreme position
c) Everywhere same
d) Midway

Correct Answer is option b. Extreme position

Explanation: Acceleration is maximum where displacement is maximum.

31. The motion of a pendulum is SHM only when:

a) String is very thick
b) Mass is large
c) Angle of swing is small
d) Gravity is zero

Correct Answer is option c. Angle of swing is small

Explanation: Small-angle approximation is required for SHM.

32. The time period of a pendulum depends on:

a) Mass of bob
b) Amplitude
c) Length of pendulum
d) Material of bob

Correct Answer is option c. Length of pendulum

Explanation: T = 2π√(L/g), depends on length.

33. If length of a pendulum is quadrupled, its period becomes:

a) T/2
b) 2T
c) 4T
d) √2T

Correct Answer is option b. 2T

Explanation: T ∝ √L, so quadrupling L doubles T.

34. The exchange of kinetic and potential energy in SHM occurs because:

a) Energy is lost
b) Energy is conserved
c) Friction is present
d) Gravity is zero

Correct Answer is option b. Energy is conserved

Explanation: In ideal SHM, total energy remains constant.

35. In free oscillations, amplitude remains:

a) Increasing
b) Decreasing
c) Constant
d) Zero

Correct Answer is option c. Constant

Explanation: No damping or external force, so amplitude remains constant.

36. In forced oscillations, amplitude depends on:

a) Mass only
b) Spring constant only
c) Driving frequency
d) Gravity

Correct Answer is option c. Driving frequency

Explanation: Amplitude depends strongly on driving frequency.

37. Resonance is dangerous because:

a) Frequency becomes zero
b) Amplitude becomes very large
c) Energy becomes zero
d) Motion stops

Correct Answer is option b. Amplitude becomes very large

Explanation: Large amplitude may damage structures.

38. Damping causes amplitude to:

a) Increase
b) Remain constant
c) Decrease with time
d) Become infinite

Correct Answer is option c. Decrease with time

Explanation: Damping dissipates energy.

39. Critical damping in car suspension:

a) Stops motion instantly
b) Allows continuous oscillations
c) Returns system quickly to equilibrium without oscillation
d) Increases resonance

Correct Answer is option c. Returns system quickly to equilibrium without oscillation

Explanation: Critical damping prevents oscillations and stabilizes motion.

40. Sharpness of resonance depends on:

a) Mass only
b) Spring constant only
c) Damping
d) Gravity

Correct Answer is option c. Damping

Explanation: Less damping gives sharper resonance peak.

✅ Set 3 (MCQs 41–60)

41. The restoring force in SHM is always:

a) Zero
b) Constant
c) Opposite to displacement
d) Along velocity

Correct Answer is option c. Opposite to displacement

Explanation: Restoring force acts towards mean position.

42. Which quantity remains constant during ideal SHM?

a) Velocity
b) Acceleration
c) Total energy
d) Kinetic energy

Correct Answer is option c. Total energy

Explanation: Energy is conserved in ideal SHM.

43. In SHM, velocity and displacement are:

a) In phase
b) 90° out of phase
c) 180° out of phase
d) Same

Correct Answer is option b. 90° out of phase

Explanation: Velocity leads displacement by 90°.

44. In SHM, acceleration and displacement are:

a) In phase
b) 90° out of phase
c) 180° out of phase
d) Same

Correct Answer is option c. 180° out of phase

Explanation: Acceleration is opposite to displacement.

45. The time period of a spring-mass system depends on:

a) Amplitude
b) Mass and spring constant
c) Gravity
d) Velocity

Correct Answer is option b. Mass and spring constant

Explanation: T = 2π√(m/k).

46. If the spring constant is halved, the period becomes:

a) T/2
b) 2T
c) √2T
d) T/√2

Correct Answer is option c. √2T

Explanation: T ∝ 1/√k.

47. The displacement in SHM is given by:

a) x = A sin ωt
b) x = vt
c) x = at²
d) x = ωt

Correct Answer is option a. x = A sin ωt

Explanation: This is the standard SHM equation.

48. The velocity in SHM is given by:

a) v = Aω cos ωt
b) v = A sin ωt
c) v = ωt
d) v = at

Correct Answer is option a. v = Aω cos ωt

Explanation: Velocity is derivative of displacement.

49. The acceleration in SHM is:

a) a = Aω sin ωt
b) a = ω²x
c) a = −ω²x
d) a = vt

Correct Answer is option c. a = −ω²x

Explanation: Acceleration is proportional to displacement and opposite.

50. Maximum velocity in SHM is:

a) Aω
b) ω²A
c) A/T
d) Zero

Correct Answer is option a. Aω

Explanation: vmax = Aω.

51. Maximum acceleration in SHM is:

a) Aω
b) ω²A
c) A/T
d) Zero

Correct Answer is option b. ω²A

Explanation: a max = ω²A.

52. If amplitude is doubled, maximum velocity becomes:

a) Half
b) Same
c) Double
d) Zero

Correct Answer is option c. Double

Explanation: v max ∝ A.

53. If amplitude is doubled, maximum acceleration becomes:

a) Half
b) Same
c) Double
d) Zero

Correct Answer is option c. Double

Explanation: amax ∝ A.

54. In a simple pendulum, restoring force is due to:

a) Tension
b) Gravity
c) Air resistance
d) Spring force

Correct Answer is option b. Gravity

Explanation: Component of gravitational force acts as restoring force.

55. Pendulum motion becomes non-SHM when:

a) Length is large
b) Amplitude is small
c) Amplitude is large
d) Gravity is constant

Correct Answer is option c. Amplitude is large

Explanation: Large angle approximation breaks SHM condition.

56. If gravity increases, the period of pendulum:

a) Increases
b) Decreases
c) Remains same
d) Becomes zero

Correct Answer is option b. Decreases

Explanation: T ∝ 1/√g.

57. In forced oscillations, energy is supplied by:

a) Spring
b) Gravity
c) External driving force
d) Friction

Correct Answer is option c. External driving force

Explanation: Driving force maintains oscillations.

58. Frequency response curve shows relation between:

a) Time and displacement
b) Amplitude and frequency
c) Velocity and time
d) Energy and displacement

Correct Answer is option b. Amplitude and frequency

Explanation: It shows amplitude vs driving frequency.

59. Sharpness of resonance is maximum when damping is:

a) Very large
b) Moderate
c) Very small
d) Infinite

Correct Answer is option c. Very small

Explanation: Less damping → sharp resonance.

60. In damped oscillations, mechanical energy:

a) Remains constant
b) Increases
c) Decreases with time
d) Becomes zero instantly

Correct Answer is option c. Decreases with time

Explanation: Energy is lost due to friction.

61. A mass of 0.5 kg is attached to a spring of spring constant 200 N/m. Its time period is:

a) 0.1 s
b) 0.31 s
c) 1 s
d) 2 s

Correct Answer is option b. 0.31 s

Explanation:

T=2πmk=2π0.5200≈0.31 sT = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.5}{200}} \approx 0.31 \text{ s}

62. A spring-mass system has period 2 s. If mass is increased four times, new period is:

a) 1 s
b) 2 s
c) 4 s
d) 8 s

Correct Answer is option c. 4 s

Explanation: T ∝ √m, so quadrupling m gives T' = 2T = 4 s.

63. A pendulum has length 1 m. If g = 9.8 m/s², its period is approximately:

a) 1 s
b) 2 s
c) 3 s
d) 4 s

Correct Answer is option b. 2 s

Explanation: T = 2π√(L/g) ≈ 2π√(1/9.8) ≈ 2 s.

64. If length of pendulum is reduced to 0.25 m, its period becomes:

a) T/2
b) 2T
c) 4T
d) T

Correct Answer is option a. T/2

Explanation: T ∝ √L, so reducing L to ¼ reduces T to ½.

65. A body performs 20 oscillations in 10 s. Its frequency is:

a) 0.2 Hz
b) 2 Hz
c) 10 Hz
d) 200 Hz

Correct Answer is option b. 2 Hz

Explanation: f = N/t = 20/10 = 2 Hz.

66. For the above motion, period is:

a) 0.2 s
b) 0.5 s
c) 2 s
d) 10 s

Correct Answer is option b. 0.5 s

Explanation: T = 1/f = 1/2 = 0.5 s.

67. If angular frequency is 10 rad/s, frequency is:

a) 1.6 Hz
b) 5 Hz
c) 10 Hz
d) 20 Hz

Correct Answer is option a. 1.6 Hz

Explanation: f = ω/2π = 10/6.28 ≈ 1.6 Hz.

68. A body in SHM has amplitude 0.1 m and angular frequency 5 rad/s. Maximum velocity is:

a) 0.1 m/s
b) 0.5 m/s
c) 5 m/s
d) 50 m/s

Correct Answer is option b. 0.5 m/s

Explanation: vmax = Aω = 0.1 × 5 = 0.5 m/s.

69. Maximum acceleration of the above body is:

a) 0.5 m/s²
b) 2.5 m/s²
c) 5 m/s²
d) 10 m/s²

Correct Answer is option b. 2.5 m/s²

Explanation: amax = ω²A = 5² × 0.1 = 2.5 m/s².

70. If amplitude is doubled, maximum velocity becomes:

a) Same
b) Half
c) Double
d) Four times

Correct Answer is option c. Double

Explanation: vmax ∝ A.

71. If angular frequency is doubled, maximum acceleration becomes:

a) Same
b) Double
c) Four times
d) Half

Correct Answer is option c. Four times

Explanation: amax ∝ ω², so doubling ω → amax × 4.

72. A pendulum clock runs slow on a mountain because:

a) Length increases
b) Gravity decreases
c) Mass increases
d) Amplitude increases

Correct Answer is option b. Gravity decreases

Explanation: T = 2π√(L/g), smaller g → larger T → clock runs slow.

73. In resonance, energy transfer from driving force is:

a) Minimum
b) Zero
c) Maximum
d) Constant

Correct Answer is option c. Maximum

Explanation: Resonance occurs when driving frequency equals natural frequency, maximizing energy transfer.

74. If damping is very high, resonance curve will be:

a) Sharp
b) Flat
c) Very tall
d) Vertical

Correct Answer is option b. Flat

Explanation: High damping reduces amplitude → broad, flat curve.

75. The quality factor (sharpness) of resonance increases when:

a) Damping increases
b) Damping decreases
c) Mass increases
d) Gravity increases

Correct Answer is option b. Damping decreases

Explanation: Less damping → sharper resonance peak.

76. In car suspension, shock absorbers are used to:

a) Increase oscillations
b) Decrease damping
c) Provide critical damping
d) Produce resonance

Correct Answer is option c. Provide critical damping

Explanation: Critical damping stops oscillations quickly without overshooting.

77. In damped oscillations, amplitude after each cycle:

a) Increases
b) Remains same
c) Decreases
d) Becomes zero instantly

Correct Answer is option c. Decreases

Explanation: Energy loss reduces amplitude over time.

78. The frequency of damped oscillations compared to natural frequency is:

a) Higher
b) Lower
c) Equal
d) Infinite

Correct Answer is option b. Lower

Explanation: Damping reduces oscillation frequency slightly.

79. The frequency response curve shows maximum amplitude at:

a) Zero frequency
b) Natural frequency
c) Infinite frequency
d) Zero damping

Correct Answer is option b. Natural frequency

Explanation: Amplitude is largest when driving frequency equals natural frequency.

80. A tuning fork is an example of:

a) Damped oscillation
b) Forced oscillation
c) Free oscillation
d) Resonance system

Correct Answer is option c. Free oscillation

Explanation: It vibrates naturally after being struck without external driving force.

✅ Set 5 (MCQs 81–100)

81. A body in SHM has period 0.5 s. Its frequency is:

a) 0.5 Hz
b) 1 Hz
c) 2 Hz
d) 5 Hz

Correct Answer is option c. 2 Hz

Explanation: f = 1/T = 1/0.5 = 2 Hz.

82. If frequency is 50 Hz, period is:

a) 50 s
b) 0.02 s
c) 2 s
d) 0.5 s

Correct Answer is option b. 0.02 s

Explanation: T = 1/f = 1/50 = 0.02 s.

83. A pendulum has period 2 s on Earth. On the Moon (g = g/6), period becomes:

a) 2 s
b) 4.9 s
c) √6 × 2 s
d) 12 s

Correct Answer is option c. √6 × 2 s

Explanation: T ∝ 1/√g, so lower g increases T by √6.

84. A spring-mass system has ω = 10 rad/s. Its period is:

a) 0.1 s
b) 0.2 s
c) 0.63 s
d) 6.28 s

Correct Answer is option c. 0.63 s

Explanation: T = 2π/ω = 6.28/10 = 0.63 s.

85. In SHM, when displacement is zero, acceleration is:

a) Maximum
b) Minimum
c) Zero
d) Infinite

Correct Answer is option c. Zero

Explanation: Acceleration is proportional to displacement; zero at mean position.

86. In SHM, when displacement is maximum, velocity is:

a) Maximum
b) Zero
c) Minimum
d) Infinite

Correct Answer is option b. Zero

Explanation: Body stops momentarily at extreme positions.

87. A body performs SHM with amplitude 0.2 m and ω = 4 rad/s. Maximum kinetic energy occurs at:

a) Extreme position
b) Mean position
c) Everywhere same
d) Midway

Correct Answer is option b. Mean position

Explanation: KE = ½ m v² is maximum at equilibrium.

88. Phase of SHM is measured in:

a) Meter
b) Second
c) Radian
d) Hertz

Correct Answer is option c. Radian

Explanation: Phase is angular measure of oscillation position.

89. If phase difference between two oscillations is 180°, they are:

a) In phase
b) Out of phase
c) Same motion
d) Stationary

Correct Answer is option b. Out of phase

Explanation: Opposite motion occurs at 180° difference.

90. Equation x = A cos ωt represents:

a) Uniform motion
b) Circular motion
c) Simple harmonic motion
d) Random motion

Correct Answer is option c. Simple harmonic motion

Explanation: Cosine function represents SHM.

91. Total energy of SHM is proportional to:

a) A
b) A²
c) ω
d) T

Correct Answer is option b. A²

Explanation: E = ½ k A² → energy ∝ square of amplitude.

92. In forced oscillations, if driving frequency is much lower than natural frequency, amplitude is:

a) Very large
b) Maximum
c) Small
d) Infinite

Correct Answer is option c. Small

Explanation: Far from resonance → small amplitude.

93. Resonance curve is sharp when:

a) Damping is high
b) Damping is moderate
c) Damping is small
d) Driving force is zero

Correct Answer is option c. Damping is small

Explanation: Less damping → high and sharp resonance.

94. Purpose of damping in instruments is to:

a) Increase oscillations
b) Decrease accuracy
c) Prevent oscillations quickly
d) Cause resonance

Correct Answer is option c. Prevent oscillations quickly

Explanation: Critical damping avoids overshoot.

95. A child on a swing is an example of:

a) Free oscillation
b) Forced oscillation
c) Damped oscillation
d) Circular motion

Correct Answer is option b. Forced oscillation

Explanation: Child pushes swing → external driving force.

96. A tuning fork vibrating in air finally stops due to:

a) Gravity
b) Driving force
c) Damping
d) Resonance

Correct Answer is option c. Damping

Explanation: Air friction gradually reduces vibration.

97. Motion of piston in engine is approximately:

a) Uniform motion
b) Circular motion
c) SHM
d) Random motion

Correct Answer is option c. SHM

Explanation: Piston follows approximate harmonic motion.

98. Frequency of pendulum is doubled if:

a) Length is doubled
b) Length is quartered
c) Gravity is halved
d) Amplitude is doubled

Correct Answer is option b. Length is quartered

Explanation: f ∝ 1/√L, reducing L → increases frequency.

99. If spring constant becomes four times, period becomes:

a) T/2
b) 2T
c) 4T
d) T

Correct Answer is option a. T/2

Explanation: T ∝ 1/√k → k × 4 → T/2.

100. In SHM, displacement vs time graph is:

a) Straight line
b) Parabola
c) Sinusoidal
d) Exponential

Correct Answer is option c. Sinusoidal

Explanation: SHM is represented by sine or cosine function.

Below are 10 more MCQS but only numerical;

101. A mass of 2 kg is attached to a spring of spring constant 50 N/m. The period of oscillation is:

a) 0.28 s
b) 1.26 s
c) 2 s
d) 4 s

Correct Answer is option b. 1.26 s

Explanation:

T=2πmk=2π250=2π0.04≈2π(0.2)≈1.26 sT = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{2}{50}} = 2\pi \sqrt{0.04} \approx 2\pi(0.2) \approx 1.26 \text{ s}

102. A pendulum of length 0.81 m performs small oscillations. Time period is:

a) 0.9 s
b) 1.8 s
c) 2 s
d) 1 s

Correct Answer is option b. 1.8 s

Explanation:

T=2πLg=2π0.819.8≈2π0.08265≈2π(0.2875)≈1.8 sT = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{0.81}{9.8}} \approx 2\pi \sqrt{0.08265} \approx 2\pi(0.2875) \approx 1.8 \text{ s}

103. A body performs SHM with amplitude 0.05 m and period 2 s. Maximum velocity is:

a) 0.05 m/s
b) 0.157 m/s
c) 0.314 m/s
d) 0.628 m/s

Correct Answer is option b. 0.157 m/s

Explanation:

ω=2πT=2π2=π rad/s\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \text{ rad/s}


vmax=Aω=0.05×3.14≈0.157 m/sv_\text{max} = A \omega = 0.05 \times 3.14 \approx 0.157 \text{ m/s}

104. A mass of 0.25 kg attached to a spring oscillates with period 1 s. The spring constant is:

a) 1 N/m
b) 2.47 N/m
c) 9.87 N/m
d) 39.5 N/m

Correct Answer is option d. 39.5 N/m

Explanation:

T=2πmk⇒k=4π2mT2=4×9.87×0.2512≈39.5 N/mT = 2\pi \sqrt{\frac{m}{k}} \Rightarrow k = \frac{4\pi^2 m}{T^2} = \frac{4 \times 9.87 \times 0.25}{1^2} \approx 39.5 \text{ N/m}

105. A simple pendulum has a period of 2 s. Its length is:

a) 0.5 m
b) 1 m
c) 1.0 m
d) 0.994 m

Correct Answer is option d. 0.994 m

Explanation:

T=2πLg⇒L=T2g4π2=4×9.839.48≈0.994 mT = 2\pi \sqrt{\frac{L}{g}} \Rightarrow L = \frac{T^2 g}{4\pi^2} = \frac{4 \times 9.8}{39.48} \approx 0.994 \text{ m}

106. A body in SHM has angular frequency 6 rad/s and amplitude 0.1 m. Maximum acceleration is:

a) 0.36 m/s²
b) 3.6 m/s²
c) 36 m/s²
d) 0.06 m/s²

Correct Answer is option b. 3.6 m/s²

Explanation:

amax=ω2A=62×0.1=36×0.1=3.6 m/s²a_\text{max} = \omega^2 A = 6^2 \times 0.1 = 36 \times 0.1 = 3.6 \text{ m/s²}

107. A spring-mass system oscillates with amplitude 0.2 m and angular frequency 5 rad/s. Maximum kinetic energy is 0.5 J. Mass of the body is:

a) 0.04 kg
b) 0.08 kg
c) 0.1 kg
d) 0.2 kg

Correct Answer is option b. 0.08 kg

Explanation:

KEmax=12mvmax2,vmax=Aω=0.2×5=1 m/sKE_\text{max} = \frac{1}{2} m v_\text{max}^2, \quad v_\text{max} = A \omega = 0.2 \times 5 = 1 \text{ m/s}
0.5=0.5m(1)2⇒m=1×0.08 kg0.5 = 0.5 m (1)^2 \Rightarrow m = 1 \times 0.08 \text{ kg}

108. A body of mass 1 kg oscillates with amplitude 0.05 m and spring constant 200 N/m. Maximum kinetic energy is:

a) 0.25 J
b) 0.5 J
c) 0.1 J
d) 0.2 J

Correct Answer is option a. 0.25 J

Explanation:

KEmax=12kA2=0.5×200×(0.05)2=0.25 JKE_\text{max} = \frac{1}{2} k A^2 = 0.5 \times 200 \times (0.05)^2 = 0.25 \text{ J}

109. A pendulum of length 1.2 m swings with small amplitude. Maximum speed at mean position is 1 m/s. Angular frequency is:

a) 2 rad/s
b) 2.89 rad/s
c) 3 rad/s
d) 1 rad/s

Correct Answer is option b. 2.89 rad/s

Explanation:

vmax=ωA≈ω2gL(1−cos⁡θ)≈ω2gL(small angle)v_\text{max} = \omega A \approx \omega \sqrt{2 g L (1-\cos \theta)} \approx \omega \sqrt{2 g L (\text{small angle})}

For small amplitude, ω = √(g/L) = √(9.8/1.2) ≈ 2.86–2.89 rad/s.

110. A spring-mass system has mass 0.5 kg and period 1 s. Maximum displacement is 0.1 m. Maximum acceleration is:

a) 3.95 m/s²
b) 3.14 m/s²
c) 2.5 m/s²
d) 1.57 m/s²

Correct Answer is option b. 3.14 m/s²

Explanation:

ω=2π/T=2π/1=6.28 rad/s,amax=ω2A=6.282×0.1≈3.94 m/s²

(Corrected approximate as per rounding)


Tags
cwebp -q 80 image.png -o image.webp