Question

Find the velocity when $KE=PE$ of a body undergoing SHM. Given, amplitude $=x_0$ and angular frequency is $\omega$. Also, how many times in a cycle is $KE=PE$ ?

• A. $\frac{\omega x_0}{\sqrt{2}},2$
• B. $\omega x_0,2$
• C. $\frac{\omega x_0}{\sqrt{2}},4$
• D. $\omega x_0,4$

Answer 1

The correct option is C $\frac{\omega x_0}{\sqrt{2}},4$

$PE=KE$
$\Rightarrow \frac{1}{2}m{\omega }^2(x_0^2-x^2)=\frac{1}{2}m{\omega }^2{x}^2$
i.e $x=\pm \frac{x_0}{\sqrt{2}}$

At $x=\pm \frac{x_0}{\sqrt{2}}$,
velocity, $v=\omega \sqrt{x_0^2-x^2}=\frac{\omega x_0}{\sqrt{2}}$
This will be achieved 4 times in a cycle.