Question

A body executing simple harmonic motion has a maximum acceleration equal to $24m/s^2$ and maximum velocity equal to $16m/s$. The amplitude of SHM is :

• A. $\frac{64}{9}m$
• B. $\frac{1024}{9}m$
• C. $\frac{3}{32}m$
• D. $\frac{32}{9}m$

Answer 1

Answer: (D)

When particle passes through equilibrium position, velocity is maximum and acceleration is maximum at maximum displacements The velocity of a particle in SHM changes with displacement y of the particle;
When $y=0$, velocity is maximum given by
$v_{\max }=a\omega$ ... (i)
But acceleration is maximum, when $y=a$
$\therefore {\alpha }_{\max }={\omega }^2\alpha$ ...(ii)
where a is amplitude.
Squaring Eq. (i) and dividing by Eq. (ii), we get
$\frac{v_{\max }^2}{{\alpha }_{\max }}=a$.
Given, $v_{\max }=16m/s,{\alpha }_{\max }=24m/s^2$
$\therefore a=\frac{(16{)}^2}{24}=\frac{32}{3}m$