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JEE Main 2026 (28 Jan, Shift 1) · Energy in SHM

Energy in SHM question from JEE Main 2026 (28 Jan, Shift 1)

Q17JEE Main 2026 · 28 JAN · SHIFT 1Medium

Energy in SHM

The displacement of a particle, executing simple harmonic motion with time period $T$ , is expressed as $x(t) = A\sin\omega t$ , where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T/2\beta$ . The value of $\beta$ is ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

$U = \frac{1}{2}kA^2\sin^2\omega t$ is maximum when $\sin\omega t = 1$ , i.e. $\omega t = \frac{\pi}{2}$ .
First time: $t = \frac{\pi/2}{2\pi/T} = \frac{T}{4}$
$\frac{T}{2\beta} = \frac{T}{4} \Rightarrow \beta = 2$
So $\beta = 2$ .

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