The velocity of a | SIMPLE HARMONIC MOTION | JEE MAIN_2026

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04-04-26_S2

Question

The velocity of a particle executing simple harmonic motion along $x$-axis is described as $v^2=50-x^2$, where x represents displacement. If the time period of motion is $\frac{x}{7} \mathrm{~s}$, the value of x is $\_\_\_\_$ . $(\pi=22 / 7)$

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Solution

$$ \mathrm{V}^2=\omega^2 \mathrm{~A}^2-\omega^2 \mathrm{x}^2 $$ Comparing $\omega=1$ $$ \begin{aligned} & \mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{1}=2 \pi=\frac{\mathrm{x}}{7} \\ & \mathrm{x}=14 \pi \\ & \mathrm{x}=44 \end{aligned} $$

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