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MCQEasyJEE 2025Simple Pendulum

JEE Physics 2025 Question with Solution

Two simple pendulums having lengths l1l_{1} and l2l_{2} with negligible string mass undergo angular displacements θ1\theta_{1} and θ2\theta_{2}, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?

  • A

    θ1l22=θ2l12\theta_{1} l_{2}^{2}=\theta_{2} l_{1}^{2}

  • B

    θ1l1=θ2l2\theta_{1} l_{1}=\theta_{2} l_{2}

  • C

    θ1l12=θ2l22\theta_{1} l_{1}^{2}=\theta_{2} l_{2}^{2}

  • D

    θ1l2=θ2l1\theta_{1} l_{2}=\theta_{2} l_{1}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Two simple pendulums have lengths l1l_1 and l2l_2, and angular displacements θ1\theta_1 and θ2\theta_2. Their angular accelerations are the same.

Find: The correct relation between θ1,θ2,l1,\theta_1, \theta_2, l_1, and l2l_2.

For a simple pendulum in small oscillation,

α=−ω2θ\alpha = -\omega^2 \theta

and

ω=gl\omega = \sqrt{\frac{g}{l}}

So,

α=−glθ\alpha = -\frac{g}{l}\theta

For the two pendulums,

α1=−gl1θ1,α2=−gl2θ2\alpha_1 = -\frac{g}{l_1}\theta_1, \qquad \alpha_2 = -\frac{g}{l_2}\theta_2

Since the angular accelerations are same,

α1=α2\alpha_1 = \alpha_2

Therefore,

−gl1θ1=−gl2θ2-\frac{g}{l_1}\theta_1 = -\frac{g}{l_2}\theta_2

Canceling the negative sign and gg,

θ1l1=θ2l2\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2}

Hence,

θ1l2=θ2l1\theta_1 l_2 = \theta_2 l_1

Therefore, the correct option is D.

Using restoring acceleration of a pendulum

Given: Both pendulums have equal angular acceleration.

Find: Which option matches the displacement-length relation.

Concept used: For small angular displacement, sin⁡θ≈θ\sin\theta \approx \theta, so the angular acceleration of a simple pendulum is

α=−glsin⁡θ≈−glθ\alpha = -\frac{g}{l}\sin\theta \approx -\frac{g}{l}\theta

Now write for each pendulum:

α1=−gl1θ1\alpha_1 = -\frac{g}{l_1}\theta_1 α2=−gl2θ2\alpha_2 = -\frac{g}{l_2}\theta_2

Given that both are equal in magnitude,

α1=α2\alpha_1 = \alpha_2

So,

−gl1θ1=−gl2θ2-\frac{g}{l_1}\theta_1 = -\frac{g}{l_2}\theta_2

This gives

θ1l1=θ2l2\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2}

Cross-multiplying,

θ1l2=θ2l1\theta_1 l_2 = \theta_2 l_1

This matches option D.

The correct relation is θ1l2=θ2l1\theta_1 l_2 = \theta_2 l_1.

Common mistakes

  • Using the time period formula instead of angular acceleration directly. The question asks about equality of angular accelerations, so use α=−glθ\alpha = -\frac{g}{l}\theta for small oscillations, not only T=2πlgT = 2\pi\sqrt{\frac{l}{g}}.

  • Forgetting the small-angle approximation. The relation α=−glθ\alpha = -\frac{g}{l}\theta is valid when sin⁡θ≈θ\sin\theta \approx \theta in radians. Do not use degrees or large-angle motion here.

  • Cross-multiplying incorrectly after writing θ1l1=θ2l2\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2}. The correct result is θ1l2=θ2l1\theta_1 l_2 = \theta_2 l_1, not θ1l1=θ2l2\theta_1 l_1 = \theta_2 l_2.

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