Question

Velocity of a particle varies with its displacement as $v=\left(\sqrt{9-x^2}\right)$ m/s
Find the magnitude of maximum acceleration of the particle.

Answer 1

Comparing the given equation with standard velocity-displacement
equation of simple harmonic motion, i.e, $v=\omega \sqrt{A^2-x^2},$ we get
$v=\omega$ = 1rad/s and A = 3 m
The magnitude of maximum acceleration of the particle in SHM =$v={\omega }^{2}A$
$={\left(1\right)}^2\left(3\right)m/s^2=3m/s^2$