The velocity of a particle executing SHM vari | vedclass

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(C) The given equation is $4v^2 = 50 - x^2$.Dividing by $4$,we get $v^2 = \frac{50}{4} - \frac{x^2}{4} = 12.5 - \frac{x^2}{4}$.Comparing this with the

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$88$

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(C) The given equation is $4v^2 = 50 - x^2$.Dividing by $4$,we get $v^2 = \frac{50}{4} - \frac{x^2}{4} = 12.5 - \frac{x^2}{4}$.Comparing this with the standard $SHM$ velocity equation $v^2 = \omega^2(A^2 - x^2)$,we rewrite our equation as $v^2 = \frac{1}{4}(50 - x^2) = \frac{50}{4}(1 - \frac{x^2}{50})$.Thus,$\omega^2 = \frac{1}{4}$,which gives $\omega = \frac{1}{2} \ rad/s$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{1/2} = 4\pi$.Given $\pi = \frac{22}{7}$,we have $T = 4 	imes \frac{22}{7} = \frac{88}{7} \ s$.Comparing this with the given time period $\frac{x}{7} \ s$,we find $x = 88$.

The velocity of a particle executing $SHM$ varies with displacement $(x)$ as $4v^2 = 50 - x^2$. The time period of oscillations is $\frac{x}{7} \ s$. The value of $x$ is $............$ (Take $\pi = \frac{22}{7}$)

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