A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then :

Simple Harmonic Motion PYQ — JEE Main 2019 · 11 Jan · Shift 2 · Q49 with Solution

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Correct answer: B

Use SHM form for a pendulum (small oscillations)

For a simple pendulum of length $l$ , the angular frequency is

\omega = \sqrt{\frac{g}{l}}.

If the bob executes SHM with linear amplitude $A$ , then the maximum speed is

v_{\max} = \omega A.

Hence maximum kinetic energy is

K_{\max} = \frac12 m v_{\max}^2 = \frac12 m \omega^2 A^2.

First pendulum

For length $l$ ,

\omega_1^2 = \frac{g}{l}.

So,

K_1 = \frac12 m \omega_1^2 A^2 = \frac12 m \frac{g}{l} A^2.

Second pendulum: length doubled

Now length becomes $2l$ . Then

\omega_2^2 = \frac{g}{2l}.

With the same linear amplitude $A$ ,

K_2 = \frac12 m \omega_2^2 A^2 = \frac12 m \frac{g}{2l} A^2.

Thus,

K_2 = \frac12 \left(\frac12 m \frac{g}{l} A^2\right) = \frac{K_1}{2}.

Check options

A: $K_2 = \dfrac{K_1}{2}$ ✅
B: $K_2 = 2K_1$ ❌
C: $K_2 = K_1$ ❌
D: $K_2 = \dfrac{K_1}{4}$ ❌

Remark on possible ambiguity

If “same amplitude” meant same angular amplitude $\theta_0$ , then linear amplitude would double when length doubles, and one would get

K_{\max} = \frac12 m g l \theta_0^2,

so doubling $l$ would give $K_2 = 2K_1$ .

However, in SHM notation, amplitude is generally taken as the displacement variable of the motion. For a pendulum treated as SHM along the arc, that means same linear amplitude, giving

K_2 = \frac{K_1}{2}.

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