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(B) The maximum velocity of a particle in $SHM$ is given by $v = a \omega$,where $a$ is the amplitude of displacement and $\omega$ is the angular freq

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$\frac{v^{2}}{a}$

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(B) The maximum velocity of a particle in $SHM$ is given by $v = a \omega$,where $a$ is the amplitude of displacement and $\omega$ is the angular frequency.From this,we can find the angular frequency: $\omega = \frac{v}{a}$.The maximum acceleration (amplitude of acceleration) is given by $A_{max} = \omega^2 a$.Substituting the value of $\omega$ into the acceleration formula:$A_{max} = \left(\frac{v}{a}ight)^2 	imes a = \frac{v^2}{a^2} 	imes a = \frac{v^2}{a}$.

For a particle in $SHM$,if the amplitude of the displacement is $a$ and the amplitude of velocity is $v$,the amplitude of acceleration is

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