Question

What is the maximum acceleration of the particle doing the SHM?
$y=2\sin [\frac{\pi t}{2}+\varphi ]$, 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

The correct option is B $\frac{{\pi }^2}{2}$ cm/s${}^2$

$y=2\sin [\frac{\pi t}{2}+\varphi ]$, where $2$ is in cm
Compare the equation with the standard equation,

$A=\sin (wt+\varphi )$

$\Rightarrow A=2cm,w=\frac{\pi }{2}$

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${}^2$