Question

A body executing simple harmonic motion has a maximum acceleration equal to $ 24\,m/s^{2}$ and maximum velocity equal to $16\, m/s$. 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
$\frac{v^{2}_{\max }}{\alpha_{\max}}=a$.
Given, $v_{\max}=16\,m/s,\,\alpha_{\max}=24\,m/s^{2}$
$\therefore a=\frac{(16)^{2}}{24}=\frac{32}{3}\,m$