A simple harmonic oscillator has an… - JEE Main 2024 Physics PYQ

JEE Main 2024 — Physics Waves & Oscillations

2024integermedium

A simple harmonic oscillator has an amplitude $A$ and time period $6\pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ will be $\frac{\pi }{x}s$ , where $x=_______.$

Official previous-year question

Held on 29 Jan 2024 · Verified 6 Jul 2026.

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Solution

JEE Main 2024 Physics, Oscillations — solution figure

From phasor diagram particle has to move from $P$ to $Q$ in a circle of radius equal to amplitude of $\mathrm{SHM}$ .

$\mathrm{cos}\phi =\frac{\frac{\sqrt{3}A}{2}}{A}=\frac{\sqrt{3}}{2}$

$\Rightarrow \phi =\frac{\pi }{6}$

Now, $\frac{\pi }{6}=\omega t$

$\Rightarrow \frac{\pi }{6}=\frac{2\pi }{T}t$

$\Rightarrow \frac{\pi }{6}=\frac{2\pi }{6\pi }t$

$\Rightarrow t=\frac{\pi }{2}$

So, $x=2$

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