Physics · Simple Harmonic Motion

JEE Main 2024 — 29 January, Shift 2 — Question 53

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 $\frac{\pi}{\mathrm{x}} \mathrm{s}$ , where $\mathrm{x}=$ $\qquad$ ：

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

From phasor diagram particle has to move from P to Q in a circle of radius equal to amplitude of SHM. $\cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2}$

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

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

$\frac{\pi}{6}=\frac{2 \pi}{\mathrm{~T}} \mathrm{t}$

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

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

So, $\mathrm{x}=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Kinematics of SHM, Phase and Energy in SHM