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Simple pendulum and time-period expression: NEET MCQs with answers

56 questions on this topic of Oscillations and Waves (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A simple pendulum has length 1 m. Taking g = π² m/s², its time period is:

  1. 3 s
  2. 4 s
  3. 1 s
  4. 2 s
Show answer and explanation

Answer: (D) 2 s

For a simple pendulum T = 2π√(L/g). With g = π², T = 2π√(1/π²) = 2√1 = 2 s.

medium

2. Two simple pendulums of lengths 64 cm and 49 cm start oscillating together in the same phase from their mean positions. The minimum number of oscillations completed by the shorter pendulum when they are next in phase at the mean position is:

  1. 7
  2. 14
  3. 9
  4. 8
Show answer and explanation

Answer: (D) 8

The ratio of their time periods is T₁ / T₂ = √(L₁ / L₂) = √(64 / 49) = 8 / 7. Let n₁ and n₂ be the number of oscillations completed by the longer and shorter pendulums respectively when they come into the same phase again. Then n₁ T₁ = n₂ T₂, which gives n₁ / n₂ = T₂ / T₁ = 7 / 8. Thus, the shorter pendulum completes n₂ = 8 oscillations.

hard

3. A simple pendulum of length L is suspended from the ceiling. A horizontal nail is fixed at a distance of 3L/4 vertically below the point of suspension. The bob is released from a small angle. The time period of this system is:

  1. (5/4) π √(L/g)
  2. 2π √(L/g)
  3. (3/2) π √(L/g)
  4. 3π √(L/g)
Show answer and explanation

Answer: (C) (3/2) π √(L/g)

For one half of the oscillation (say, to the left), the pendulum swings freely with length L, taking time t₁ = (1/2) × 2π√(L/g) = π√(L/g). For the other half (to the right), the string bends around the nail, so effective length is L - 3L/4 = L/4, taking time t₂ = (1/2) × 2π√(L/(4g)) = (π/2)√(L/g). Total period T = t₁ + t₂ = π√(L/g) + (π/2)√(L/g) = (3/2)π√(L/g).

medium

4. A simple pendulum has time period T. If it is placed inside a spacecraft orbiting Earth in a circular orbit at a height h = R (where R is the Earth's radius), its time period will be:

  1. 2 T
  2. 4 T
  3. Infinite
  4. T / 2
Show answer and explanation

Answer: (C) Infinite

Inside an orbiting spacecraft, all objects are in a state of free fall, so the effective acceleration due to gravity is g_eff = 0. Therefore, the time period T = 2π√(L/g_eff) becomes infinite.

easy

5. The length of a simple pendulum is decreased by 36%. The percentage decrease in its time period of oscillation for small amplitudes is:

  1. 18%
  2. 64%
  3. 36%
  4. 20%
Show answer and explanation

Answer: (D) 20%

The time period is related to length by T ∝ √L. If the new length is L' = L - 0.36L = 0.64L, then the new time period is T' = √(0.64) T = 0.8 T. The percentage decrease in time period is ((T - T') / T) × 100% = (1 - 0.8) × 100% = 20%.

hard

6. A clock regulated by a simple pendulum keeps correct time on the Earth's surface. If it is taken to a height equal to the radius of Earth (h = R), how much time will it lose per day (24 hours)?

  1. 12 hours
  2. 6 hours
  3. 18 hours
  4. 24 hours
Show answer and explanation

Answer: (A) 12 hours

At height h = R, the acceleration due to gravity is g' = g(R/(R+R))² = g/4. The time period becomes T' = 2π√(L/g') = 2T. Since the pendulum takes twice as long per oscillation, the clock runs at half the normal rate, showing only 12 hours in a 24-hour day and losing 12 hours.

medium

7. Two simple pendulums of lengths 1.44 m and 1.00 m start swinging simultaneously in the same phase from their mean position. After how many oscillations of the shorter pendulum will they again be in the same phase at the mean position?

  1. 5
  2. 6
  3. 12
  4. 10
Show answer and explanation

Answer: (B) 6

The time periods are T₁ = 2π√(1.44/g) ∝ 1.2 and T₂ = 2π√(1.00/g) ∝ 1.0. Thus, T₁/T₂ = 1.2/1.0 = 6/5. For them to be in phase again, n₂ T₂ = n₁ T₁, which gives n₂/n₁ = T₁/T₂ = 6/5. Hence, the shorter pendulum completes 6 oscillations when the longer completes 5.

easy

8. A simple pendulum has length 0.49 m. Taking g = π² m/s², its time period is:

  1. 0.98 s
  2. 1.4 s
  3. 0.7 s
  4. 2.8 s
Show answer and explanation

Answer: (B) 1.4 s

For a simple pendulum T = 2π√(L/g). With g = π², T = 2π√(0.49/π²) = 2√0.49 = 1.4 s.

easy

9. Consider the following two statements regarding a simple pendulum: Statement I: The restoring torque acting on the pendulum bob is directly proportional to sin θ, where θ is the angular displacement from the vertical. Statement II: The motion of a simple pendulum is strictly simple harmonic only in the limit of small angular displacement. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

The restoring torque is τ = -mgL sin θ. Since restoring torque is proportional to sin θ rather than θ, the motion is strictly simple harmonic only when sin θ ≈ θ, which holds true for small angular displacements.

easy

10. A simple pendulum has length 4 m. Taking g = π² m/s², its time period is:

  1. 6 s
  2. 4 s
  3. 2 s
  4. 8 s
Show answer and explanation

Answer: (B) 4 s

For a simple pendulum T = 2π√(L/g). With g = π², T = 2π√(4/π²) = 2√4 = 4 s.

medium

11. Consider the following two statements regarding a simple pendulum: Statement I: The motion of a simple pendulum is strictly simple harmonic for all angular displacements θ up to 90°. Statement II: The restoring torque is proportional to sin θ, which can be approximated as θ only for small values of θ. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (C) Statement I is incorrect but Statement II is correct

The restoring torque is τ = -mgL sin θ. For SHM, the restoring torque must be directly proportional to θ (τ ∝ -θ), which holds only when sin θ ≈ θ (valid for small angles). Thus, Statement I is incorrect and Statement II is correct.

medium

12. A brass wire supports the bob of a simple pendulum. When the temperature increases by Δθ = 20 °C, the fractional increase in the time period of the pendulum is: (Coefficient of linear expansion of brass α = 2.0 × 10⁻⁵ /°C)

  1. 2.0 × 10⁻⁴
  2. 4.0 × 10⁻⁴
  3. 1.0 × 10⁻⁴
  4. 8.0 × 10⁻⁴
Show answer and explanation

Answer: (A) 2.0 × 10⁻⁴

The fractional change in length is ΔL/L = α Δθ. Since T = 2π√(L/g), the fractional change in time period is ΔT/T = (1/2)(ΔL/L) = (1/2) α Δθ = (1/2)(2.0 × 10⁻⁵)(20) = 2.0 × 10⁻⁴.

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