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Simple pendulum: energy dissipation graph of amplitude squared versus time: NEET MCQs with answers

50 questions on this topic of Experimental Skills (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In an experiment to plot the dissipation curve, the amplitude of a simple pendulum decreases by 10% after 20 oscillations. The percentage decrease in its mechanical energy during this interval is:

  1. 10%
  2. 19%
  3. 20%
  4. 81%
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Answer: (B) 19%

The energy is proportional to the square of amplitude (E ∝ A²). If amplitude decreases by 10%, the new amplitude is A' = 0.9 A₀. Therefore, the remaining energy is E' = (0.9)² E₀ = 0.81 E₀, which corresponds to a decrease of (1 - 0.81) × 100% = 19%.

medium

2. Consider the following two statements regarding the energy dissipation graph of a simple pendulum: Statement I: The graph of amplitude squared (A²) versus time (t) is concave upwards. Statement II: The instantaneous rate of dissipation of mechanical energy (|dE/dt|) is constant throughout the motion.

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

Since A² decays exponentially (A² = A₀² e^(–2bt)), the second derivative d²(A²)/dt² = 4b²A₀² e^(–2bt) > 0, which means the curve is concave upwards. The rate of dissipation |dE/dt| = 2bE is proportional to energy, so it decreases continuously over time rather than remaining constant.

hard

3. Two identical simple pendulums 1 and 2 start oscillating with equal initial amplitude A₀. Pendulum 1 oscillates in air while Pendulum 2 oscillates in a light oil chamber. At a certain time t₁, the amplitude of pendulum 1 is A₀/2 and that of pendulum 2 is A₀/4. The ratio of energy dissipated by pendulum 2 to that by pendulum 1 up to time t₁ is:

  1. 5 : 3
  2. 4 : 5
  3. 15 : 12
  4. 5 : 4
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Answer: (D) 5 : 4

For pendulum 1, remaining energy is E₁ = E₀(A₁/A₀)² = E₀(1/4) = 0.25 E₀, so dissipated energy is ΔE₁ = 0.75 E₀. For pendulum 2, remaining energy is E₂ = E₀(A₂/A₀)² = E₀(1/16) = 0.0625 E₀, so dissipated energy is ΔE₂ = 0.9375 E₀ = 15/16 E₀. The ratio ΔE₂ / ΔE₁ = (15/16) / (3/4) = (15/16) × (4/3) = 5/4.

medium

4. A student measures the amplitude of a simple pendulum as a function of time and records the following data: At t = 0 s, A = 8.0 cm At t = 40 s, A = 4.0 cm Assuming exponential decay, at what time will the square of amplitude (A²) become 1.0 cm²?

  1. 80 s
  2. 160 s
  3. 120 s
  4. 240 s
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Answer: (C) 120 s

At t = 0, A² = 64 cm². At t = 40 s, A = 4.0 cm, so A² = 16 cm². In 40 s, A² decreases by a factor of 4. For A² to drop from 64 cm² to 1.0 cm² is a reduction by a factor of 64 = 4³. Hence, it takes 3 intervals of 40 s, which is 3 × 40 s = 120 s.

easy

5. In the standard laboratory procedure for plotting the dissipation curve of a simple pendulum, why is the initial angular displacement kept small (less than about 5°)?

  1. To avoid dissipation of energy so that mechanical energy remains conserved
  2. To prevent air drag from acting on the oscillating bob
  3. To make the time period of oscillation depend on the amplitude of swing
  4. To ensure the motion remains strictly within the simple harmonic motion regime where restoring force is proportional to displacement
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Answer: (D) To ensure the motion remains strictly within the simple harmonic motion regime where restoring force is proportional to displacement

The analysis of a simple pendulum assumes simple harmonic motion, which requires sin θ ≈ θ (valid for small angles, typically θ < 5°). If the angle is large, the restoring torque is non-linear and the period varies with amplitude, complicating the energy dissipation relationship.

hard

6. During an experiment to plot the dissipation curve (A² versus t) of a simple pendulum, a student records the amplitude at equal intervals of time. What is the physical significance of the area under the A² versus t curve from t = 0 to t → ∞?

  1. It is directly proportional to the total initial mechanical energy divided by the damping constant
  2. It equals the total work done by air drag in one single complete cycle
  3. It represents the quality factor divided by the time period
  4. It gives the instantaneous power lost at t = 0
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Answer: (A) It is directly proportional to the total initial mechanical energy divided by the damping constant

Since total mechanical energy is E(t) = (1/2)mω²A²(t) and A²(t) = A₀² e^(–2bt), the integral ∫ A² dt from 0 to ∞ equals A₀² / (2b). Since E₀ = (1/2)mω²A₀², the integral is proportional to E₀ / b, reflecting initial energy normalized by the damping rate.

medium

7. To obtain an accurate A² versus time curve for a simple pendulum in the laboratory, why should the initial angular displacement θ₀ be kept small (typically less than 10°)?

  1. To ensure the period depends strongly on amplitude
  2. To keep the motion strictly within the simple harmonic motion approximation where sin θ ≈ θ
  3. To increase the viscous drag of air on the pendulum bob
  4. To prevent any energy from being dissipated into the surroundings
Show answer and explanation

Answer: (B) To keep the motion strictly within the simple harmonic motion approximation where sin θ ≈ θ

The theoretical relationship E = (1/2) m ω² A² relies on the pendulum executing simple harmonic motion, which is strictly valid only in the small-angle approximation (sin θ ≈ θ). For larger angles, restoring torque is non-linear and the period depends on amplitude.

medium

8. In an energy dissipation experiment, the amplitude of a pendulum decays as A(t) = A₀ e^(–λt). If it takes 40 s for A² to decrease to (1/3) of its initial value A₀², the time taken for A² to decrease to (1/9) of A₀² is:

  1. 60 s
  2. 80 s
  3. 120 s
  4. 160 s
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Answer: (B) 80 s

Since A²(t) = A₀² e^(–2λt), the quantity A² undergoes exponential decay with a constant fractional reduction per unit time. If A² reduces by a factor of 3 in 40 s, it will reduce by a factor of 3² = 9 in twice that time, which is 2 × 40 s = 80 s.

easy

9. For a simple pendulum oscillating with small amplitude A, its mechanical energy is proportional to:

  1. A²
  2. A
  3. √A
  4. 1/A
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Answer: (A) A²

The energy of an oscillator is ½mω²A², proportional to the square of the amplitude. That is why the experiment plots A² against time to study energy loss.

NCERT Class 11

medium

10. A student performs the experiment of plotting A² versus time for a simple pendulum. If the bob is replaced by another bob of the same mass and material but larger surface area, the dissipation curve will:

  1. Become a straight line parallel to the time axis
  2. Decay more slowly, sustaining oscillations longer
  3. Remain unchanged because the mass has not changed
  4. Decay more rapidly, reaching lower A² values in less time
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Answer: (D) Decay more rapidly, reaching lower A² values in less time

Air resistance depends directly on the effective cross-sectional area of the bob. A larger surface area leads to a larger damping constant b, meaning mechanical energy and consequently A² will dissipate at a faster rate.

easy

11. A simple pendulum oscillates in air with light damping. If the amplitude decays according to A(t) = A₀ e^(–γt), where γ is the damping constant, how does A² depend on time t?

  1. A² = A₀² e^(–γt)
  2. A² = A₀² (1 – 2γt)
  3. A² = A₀² e^(–2γt)
  4. A² = A₀² e^(–γt/2)
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Answer: (C) A² = A₀² e^(–2γt)

Squaring the amplitude function A(t) = A₀ e^(–γt) yields A²(t) = [A₀ e^(–γt)]² = A₀² e^(–2γt). Thus, the square of the amplitude decays exponentially with an effective decay constant of 2γ.

hard

12. In a simple pendulum energy dissipation experiment, the time taken for the amplitude squared A² to drop to 25% of its initial value is t₁. The time taken for A² to drop from 25% to 5% of the initial value is t₂. Assuming A²(t) = A₀² e^(–kt), the ratio t₂ / t₁ is:

  1. ln(5) / ln(4)
  2. ln(4) / ln(5)
  3. ln(20) / ln(4)
  4. 5 / 4
Show answer and explanation

Answer: (A) ln(5) / ln(4)

We have A²(t) = A₀² e^(–kt). At t₁, A²(t₁) = 0.25 A₀², which gives e^(–kt₁) = 0.25, so kt₁ = ln(1/0.25) = ln(4). At time (t₁ + t₂), A² = 0.05 A₀², so k(t₁ + t₂) = ln(1/0.05) = ln(20). Therefore, kt₂ = ln(20) – ln(4) = ln(20/4) = ln(5). Thus, the ratio t₂ / t₁ = ln(5) / ln(4).

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