Question

What is the maximum acceleration of the particle doing the SHM $y = 2\sin\left\lbrack \frac{\pi t}{2} + \varphi \right\rbrack$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. $\frac{\mathbf{\pi}}{\mathbf{4}}\mathbf{cm/}\mathbf{s}^{\mathbf{2}}$

Answer 1

Answer: (B)

$\frac{\pi^{2}}{2}cm/s^{2}$

Answer 2

Comparing given equation with standard equation,

$y = a\sin(\omega t + \varphi),$ we get, $a = 2cm,$ $\omega = \frac{\pi}{2}$

$\therefore A_{\max} = \omega^{2}A = \left( \frac{\pi}{2} \right)^{2} \times 2$ $= \frac{\pi^{2}}{2}cm/s^{2}$.