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Question

(C) Given,amplitude $A = 5\, cm$ and displacement $x = 4\, cm$.The magnitude of velocity in simple harmonic motion is $|v| = \omega \sqrt{A^2 - x^2}$.

Vedclass · Accepted Answer

$\frac{8\pi}{3}$

Vedclass · Answer

(C) Given,amplitude $A = 5\, cm$ and displacement $x = 4\, cm$.The magnitude of velocity in simple harmonic motion is $|v| = \omega \sqrt{A^2 - x^2}$.The magnitude of acceleration in simple harmonic motion is $|a| = \omega^2 x$.According to the problem,$|v| = |a|$ at $x = 4\, cm$:$\omega \sqrt{A^2 - x^2} = \omega^2 x$Substituting the values:$\omega \sqrt{5^2 - 4^2} = \omega^2 	imes 4$$\omega \sqrt{25 - 16} = 4\omega^2$$\omega 	imes 3 = 4\omega^2$$\omega = \frac{3}{4}\, rad/s$The periodic time $T$ is given by $T = \frac{2\pi}{\omega}$:$T = \frac{2\pi}{3/4} = \frac{8\pi}{3}\, s$.

$A$ particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position,the magnitude of its velocity is equal to the magnitude of its acceleration. Then,its periodic time in seconds is

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