Question

A particle executes $S.H.M.$ of amplitude $a$
1- At what distance from mean position is its kinetic energy equal to its potential energy
2- At what points is its speed half the maximum speed

Answer 1

$\left(1\right)$ Let x be the distance where KE is equal to PE.

Kinetic energy of particles executing SHM is $KE=\frac{1}{2}m{\omega }^2(a^2-x^2)$

Potential energy of particles is given by $PE=\frac{1}{2}m{\omega }^2{x}^2$

Le x be the distance where

$KE=PE$

$\frac{1}{2}m{\omega }^2(a^2-x^2)=\frac{1}{2}m{\omega }^2{x}^2$

$(a^2-x^2)=x^2$

$x=\frac{a}{\sqrt{2}}$

kinetic energy equal to its potential energy at $x=\frac{a}{\sqrt{2}}$.

$\left(2\right)$ velocity of particles executing SHM is given by $v^{{}^{\prime }}=\omega \sqrt{(a^2-x^2)}$

Maximum velocity $v_{max}=a\omega$

Let x be the point where the speed becomes half of the maximum speed. Then

$v^{\prime }=\frac{v_{max}}{2}$

$\omega \sqrt{(a^2-x^2)}=\frac{a\omega }{2}$

${\omega }^2(a^2-x^2)=\frac{a^2{\omega }^2}{4}$

$a^2{\omega }^2-x^2{\omega }^2=\frac{a^2{\omega }^2}{4}$

$4a^2{\omega }^2-4x^2{\omega }^2=a^2{\omega }^2$

$3a^2{\omega }^2=4x^2{\omega }^2$

$x^2=\frac{3a^2}{4}$

$x=\frac{\sqrt{3}}{2}a$

Speed becomes half of the maximum speed at $x=\frac{\sqrt{3}}{2}a$.