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Question

(B) The speed $V$ of a particle performing $S.H.M.$ at a displacement $x$ from the mean position is given by the formula: $V = \omega \sqrt{a^2 - x^2}

Vedclass · Accepted Answer

$\frac{4 \sqrt{2} \pi a}{3 T}$

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(B) The speed $V$ of a particle performing $S.H.M.$ at a displacement $x$ from the mean position is given by the formula: $V = \omega \sqrt{a^2 - x^2}$.Given,angular frequency $\omega = \frac{2 \pi}{T}$ and displacement $x = \frac{a}{3}$.Substituting these values into the formula:$V = \frac{2 \pi}{T} \sqrt{a^2 - (\frac{a}{3})^2}$$V = \frac{2 \pi}{T} \sqrt{a^2 - \frac{a^2}{9}}$$V = \frac{2 \pi}{T} \sqrt{\frac{8 a^2}{9}}$$V = \frac{2 \pi}{T} 	imes \frac{2 \sqrt{2} a}{3}$$V = \frac{4 \sqrt{2} \pi a}{3 T}$

$A$ particle is performing $S.H.M.$ about its mean position with an amplitude $a$ and periodic time $T$. The speed of the particle when its displacement from the mean position is $\frac{a}{3}$ will be:

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