Question

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

• A. $v\alpha$
• B. $\frac{v^2}{\alpha }$
• C. $\frac{v^2}{2\alpha }$
• D. $\frac{v}{\alpha }$

Answer 1

Answer: (B)

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

Since the amplitude of displaement of the SHM is $\alpha$, the SHM would have the form,
$x=\alpha sin(\omega t+\varphi )$

Thus $v=\frac{dx}{dt}=\omega \alpha cos(\omega t+\varphi )$

$⟹$ amplitude of velocity=$\omega \alpha =v$

$⟹\omega =\frac{v}{\alpha }$

Similarly, acceleration of the particle in SHM is $\frac{dv}{dt}=-{\omega }^2\alpha sin(\omega t+\varphi )$

$⟹$ amplitude of acceleartion=${\omega }^2\alpha =(\frac{v}{\alpha }{)}^2\alpha =\frac{v^2}{\alpha }$