Question

What is the maximum acceleration of the particle doing the SHM?
$y = 2 \sin \left[\frac{\pi t}{2} + \phi \right]$ , where $2$ is in cm

• A. $\frac{\pi}{2} cm /s^2$
• B. $\frac{\pi^2}{2} cm /s^2$
• C. $\frac{\pi}{4} cm /s^2$
• D. $\frac{\pi^2}{4} cm /s^2$

Answer 1

Answer: (B)

$y =2 \sin \left[\frac{\pi t }{2}+\phi\right]$,
where $2$ is in $cm$
Compare the equation with the standard equation,
$A=\sin (w t+\phi)$
$\Rightarrow A=2 \,cm ,$
$ w=\frac{\pi}{2}$
Acceleration of the particle is
$a=w^{2} x$
At $x =+ A , a = a _{\max }$
$\therefore a _{\max }= w ^{2} A =\frac{\pi^{2}}{2} \,cm / s ^{2}$