Question

A particle is executing simple harmonic motion with amplitude of $0.1m$. At a certain instant when its displacement is $0.02$ and its acceleration is $0.5m/s^2$, the maximum velocity of the particle is (in $m/s$) :

• A. $0.01$
• B. $0.05$
• C. $0.5$
• D. $0.25$

Answer 1

The correct option is C $0.5$

Maximum acceleration

$\alpha ={\omega }^{2}y$

$\therefore 0.5={\omega }^2\times 0.02$

or ${\omega }^2=\frac{0.5}{0.02}=25$

So, $\omega =5$

Now, maximum velocity is

$v=a\omega =0.1\times 5=0.5m/s$