Question

$E$ is kinetic energy of a simple harmonic oscillator at its mean position. The phase angle from mean position at which its kinetic energy is $E/2$ is :

• A. $\pi /5$ rad
• B. $\pi /4$ rad
• C. $\pi /3$ rad
• D. None of the above

Answer 1

The correct option is B $\pi /4$ rad

$v=\omega \sqrt{A^2-x^2}$
$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2(A^2-x^2)$
At $x=0;KE=E$
$\frac{1}{2}m{\omega }^2{A}^2=E$
For $KE$ to be $\frac{E}{2}$
$\frac{1}{2}\times \text{/}\frac{1}{2}\text{/}m\text{/}{\omega }^2{A}^2=\text{/}\frac{1}{2}\text{/}m\text{/}{\omega }^2(A^2-x^2)$
$\frac{A^2}{2}=A^2-x^2;x^2=\frac{A^2}{2}$
$x=\frac{A}{\sqrt{2}}$
$x$ can be written as A cos $\varphi$
When $\varphi$
is $\frac{\pi }{4}$ the x is $\frac{A}{\sqrt{2}}$