Question

For what value of displacement, the kinetic energy and potential energy of a simple harmonic oscillation become equal?

$[A$ is amplitude$]$

• A. $x=\frac{A}{2}$
• B. $x=0$
• C. $x=\pm A$
• D. $x=\pm \frac{A}{\sqrt{2}}$

Answer 1

The correct option is D $x=\pm \frac{A}{\sqrt{2}}$

Given:

$\text{KE}=\text{PE}$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

​​​​​​​$\Rightarrow \frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{1}{2}k{x}^2$

​​​​​​​$\Rightarrow \frac{1}{2}k(A^2-x^2)=\frac{1}{2}k{x}^2$

​​​​​​​​​​​​​​$\Rightarrow A^2-x^2=x^2$

​​​​​​​​​​​​​​$\Rightarrow x=\pm \frac{A}{\sqrt{2}}$