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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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $f = \frac{\omega}{2 \pi} = \frac{\sqrt{3}}{2 \pi} \, s^{-1}$.

$A \ (mm \ s^{-2})$	$16$	$8$	$0$	$-8$	$-16$
$x \ (mm)$	$-4$	$-2$	$0$	$2$	$4$

Similar Questions

Acceleration of a particle,executing $SHM$,at its mean position is

Which one of the following statements is true for the speed $v$ and the acceleration $a$ of a particle executing simple harmonic motion?

What is the maximum acceleration of a particle performing $SHM$ given by the equation $y = 2\sin \left( {\frac{{\pi t}}{2} + \phi } \right)$,where $y$ is in $cm$?

Values of the acceleration $A$ of a particle moving in simple harmonic motion as a function of its displacement $x$ are given in the table below:
$A \ (mm \ s^{-2})$ $16$ $8$ $0$ $-8$ $-16$
$x \ (mm)$ $-4$ $-2$ $0$ $2$ $4$

The period of the motion is:

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