Question

The potential energy of a simple harmonic oscillator, when the particle is half way to its end point is : (Where $E$ is the total energy)

• A. $\frac{1}{4}E$
• B. $\frac{1}{2}E$
• C. $\frac{2}{3}E$
• D. $\frac{1}{8}E$

Answer 1

The correct option is A $\frac{1}{4}E$

Potential energy of a simple harmonic oscillator
$U=\frac{1}{2}m{\omega }^2{y}^2$
Kinetic energy of a simple harmonic oscillator
$K=\frac{1}{2}m{\omega }^2(A^2-y^2)$
Here $y=$ displacement from mean position
$A=$ maximum displacement (or amplitude) from mean position
Total energy is
$E=U+K$
$=\frac{1}{2}m{\omega }^2{y}^2+\frac{1}{2}m{\omega }^2(A^2-y^2)$
$=\frac{1}{2}m{\omega }^2{A}^2$
When the particle is half way to its end point ie, at half of its amplitude then
$y=\frac{A}{2}$
Hence, potential energy
$U=\frac{1}{2}m{\omega }^2{\left(\frac{A}{2}\right)}^2$
$=\frac{1}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)$
$U=\frac{E}{4}$