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(B) Maximum velocity in $SHM$ is given by $v_{\max} = a\omega$.Maximum acceleration in $SHM$ is given by $A_{\max} = a\omega^2$,where $a$ is the ampli

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$500\sqrt{2}\,m/s^2$

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(B) Maximum velocity in $SHM$ is given by $v_{\max} = a\omega$.Maximum acceleration in $SHM$ is given by $A_{\max} = a\omega^2$,where $a$ is the amplitude and $\omega$ is the angular frequency.Given the ratio $\frac{A_{\max}}{v_{\max}} = 10$,we have $\frac{a\omega^2}{a\omega} = 10$,which implies $\omega = 10\,s^{-1}$.The displacement equation is $x = a \sin(\omega t + \phi)$.At $t = 0$,$x = 5\,m$ and $\phi = \frac{\pi}{4}$.Substituting these values: $5 = a \sin(\frac{\pi}{4}) = a \cdot \frac{1}{\sqrt{2}}$.Thus,the amplitude $a = 5\sqrt{2}\,m$.The maximum acceleration is $A_{\max} = a\omega^2 = (5\sqrt{2}) \cdot (10)^2 = 5\sqrt{2} \cdot 100 = 500\sqrt{2}\,m/s^2$.

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

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10\,s^{-1}$. At $t = 0$,the displacement is $5\,m$. What is the maximum acceleration? The initial phase is $\frac{\pi}{4}$.

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