Question

What is displacement of an object in Simple Harmonic Motion when kinetic and potential energy are equal? please explain

Answer 1

$E = E_k+E_p$

$E_k =\frac{mv^2}{2}$

$E_p =\frac{kx^2}{2}$

$\text{where x offset from equilibrium point position}$

$\text{let A oscillation amplitude,then for x = A:}$

$E=E'_k+E'_p$

$E'_k= 0 \text{ etc }v= 0$

$E_p =\frac{kA^2}{2}$

$E = \frac{kA^2}{2}$

$\text{if } E_p=E_k:$

$E_p+E_k= E;$

$2*E_p=\frac{kA^2}{2}$

$2*\frac{kx_1^2}{2}=\frac{kA^2}{2}$

$x_1= \sqrt{\frac{A^2}{2}}=0.71*A$

$\text{Answer: kinetic and potential energies are equal when shifted }$

$\text {by 0.71*A from the equilibrium point;}$

$\text{ A this is the Simple Harmonic Motion amplitude}$