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(A) The maximum velocity of a particle in $SHM$ is given by $v_{\max} = A \omega = \beta$.The maximum acceleration of a particle in $SHM$ is given by

Vedclass · Accepted Answer

$\frac{\alpha}{2 \pi \beta}$

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(A) The maximum velocity of a particle in $SHM$ is given by $v_{\max} = A \omega = \beta$.The maximum acceleration of a particle in $SHM$ is given by $a_{\max} = A \omega^2 = \alpha$.Dividing the maximum acceleration by the maximum velocity:$\frac{a_{\max}}{v_{\max}} = \frac{A \omega^2}{A \omega} = \omega$.Substituting the given values:$\frac{\alpha}{\beta} = \omega$.Since angular frequency $\omega = 2 \pi f$,where $f$ is the frequency:$\frac{\alpha}{\beta} = 2 \pi f$.Therefore,the frequency $f$ is:$f = \frac{\alpha}{2 \pi \beta}$.

Maximum velocity and maximum acceleration of a particle executing $SHM$ are $\beta$ and $\alpha$ respectively. What will be its frequency?

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