The velocity of a particle executing S.H.M. v | vedclass

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(D) Given the equation for velocity in $S.H.M.$: $4V^2 = 50 - x^2$. Rearranging this,we get $V^2 = \frac{1}{4}(50 - x^2) = \frac{1}{4}(50 - x^2)$. Com

Vedclass · Accepted Answer

$88$

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(D) Given the equation for velocity in $S.H.M.$: $4V^2 = 50 - x^2$. Rearranging this,we get $V^2 = \frac{1}{4}(50 - x^2) = \frac{1}{4}(50 - x^2)$. Comparing this with the standard $S.H.M.$ velocity equation $V^2 = \omega^2(A^2 - x^2)$,we can rewrite the given equation as $V^2 = \frac{1}{4} \cdot 50 - \frac{1}{4}x^2$. This implies $\omega^2 = \frac{1}{4}$,so $\omega = \frac{1}{2} 	ext{ rad/s}$. The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{1/2} = 4\pi$. Substituting $\pi = \frac{22}{7}$,we get $T = 4 	imes \frac{22}{7} = \frac{88}{7} 	ext{ seconds}$. Comparing this with the given time period $\frac{x}{7}$,we find $x = 88$.

The velocity of a particle executing $S.H.M.$ varies with displacement $(x)$ as $4V^2 = 50 - x^2$. The time period of oscillation is $\frac{x}{7}$ seconds. The value of '$x$' is (Take $\pi = \frac{22}{7}$)

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