Question

State an expression for $K.E$ (kinetic energy) and $P.E$(potential energy) at displacement ${}^{\prime }{x}^{\prime }$ for a particle performing linear $S.H.M$. Represent them graphically. Find the displacement at which $K.E$ is equal to $P.E$.

Answer 1

$\because KE=\frac{1}{2}m{\omega }^2(a^2-x^2)$
where $x$ is the displacement of the particle, $a$ is the amplitude.
and $PE=\frac{1}{2}m{\omega }^2{x}^2$
Now if K.E=P.E
then, $\frac{1}{2}m{\omega }^2(a^2-y^2)=\frac{1}{2}m{\omega }^2{y}^2$
$\Rightarrow \frac{1}{2}m{\omega }^2{a}^2-\frac{1}{2}m{\omega }^2{y}^2=\frac{1}{2}m{\omega }^2{y}^2$
$\Rightarrow \frac{1}{2}m{\omega }^2{a}^2=\frac{1}{2}m{\omega }^2{y}^2+\frac{1}{2}m{\omega }^2{y}^2$
$\Rightarrow \frac{1}{2}m{\omega }^2{a}^2=m{\omega }^2{y}^2$
$\Rightarrow y=\frac{a}{\sqrt{2}}$