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

Answer: (A)

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