A particle executes SHM on a straight line pa | vedclass

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(C) Given that the magnitude of velocity $|v|$ is equal to the magnitude of acceleration $|a|$.For $SHM$,velocity is $v = \omega \sqrt{A^2 - x^2}$ and

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$\frac{\sqrt{3}}{2\pi}$

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(C) Given that the magnitude of velocity $|v|$ is equal to the magnitude of acceleration $|a|$.For $SHM$,velocity is $v = \omega \sqrt{A^2 - x^2}$ and acceleration is $a = -\omega^2 x$.Equating the magnitudes: $|v| = |a| \Rightarrow \omega \sqrt{A^2 - x^2} = \omega^2 x$.Dividing both sides by $\omega$ (assuming $\omega 
eq 0$): $\sqrt{A^2 - x^2} = \omega x$.Solving for angular frequency $\omega$: $\omega = \frac{\sqrt{A^2 - x^2}}{x}$.Given $A = 2 \, cm$ and $x = 1 \, cm$,we have $\omega = \frac{\sqrt{2^2 - 1^2}}{1} = \sqrt{3} \, rad/s$.Since $\omega = 2 \pi f$,the frequency $f = \frac{\omega}{2 \pi} = \frac{\sqrt{3}}{2 \pi} \, s^{-1}$.

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