A particle executes S.H.M. of period $\frac{2 \pi}{\sqrt{3}} \sec$ along a straight line $4 \mathrm{~cm}$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is

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Aicharya · Accepted Answer

Acceleration, $a=\omega^2 x$ Velocity, $v=\omega \sqrt{A^2-x^2}$ $\begin{aligned} & \omega^2 \mathrm{x}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \ & \omega^2 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \ & 3 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \quad\left(	herefore \omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{3}ight) \ & 4 \mathrm{x}^2=\mathrm{A}^2 \ & \mathrm{x}=\frac{\mathrm{A}}{2}=\frac{2}{2}=1 \mathrm{~cm} \end{aligned}$

A particle executes S.H.M. of period $\frac{2 \pi}{\sqrt{3}} \sec$ along a straight line $4 \mathrm{~cm}$…

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