The general displacement of a simple… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Waves & Oscillations

2023integermedium

The general displacement of a simple harmonic oscillator is $x=A\mathrm{sin}\omega t$ . Let $T$ be its time period. The slope of its potential energy ( $U$ ) – time ( $t$ ) curve will be maximum when $t=\frac{T}{\beta }$ . The value of $\beta$ is ______.

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Held on 30 Jan 2023 · Verified 6 Jul 2026.

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Solution

Given here, displacement $x=A\mathrm{sin}(\omega t)$

Now, potential energy of a simple harmonic oscillator is $U=\frac{1}{2}m{\omega }^{2}{x}^{2}=\frac{1}{2}m{\omega }^{2}{A}^{2}{\mathrm{sin}}^{2}(\omega t)$ ,

So, slope of potential energy and time curve is $\frac{dU}{dt}=\frac{1}{2}\frac{m{\omega }^{3}{A}^{2}}{2}\mathrm{sin}2\omega t$

Now, ${(\frac{dU}{dt})}_{max}$ will be maximum when $\mathrm{sin}2\omega t=1$

$\Rightarrow 2\omega t=\frac{\pi }{2} \Rightarrow 2(\frac{2\pi }{T})t=\frac{\pi }{2} \Rightarrow t=\frac{T}{8} \Rightarrow \beta =8$

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