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)

$\frac{1}{4}E$

Answer 2

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 from mean position (or amplitude)
Total energy, $E=U+K$
$E =0+R$
$=\frac{1}{2} m \omega^{2} y^{2}+\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$
$=\frac{1}{2} m \omega^{2} A^{2}$
When the particle is half way to its end point ie, at half of its amplitude, then
$y=\frac{A}{2}$
Hence, potential energy
$U =\frac{1}{2} m \omega^{2}\left(\frac{A}{2}\right)^{2}$
$=\frac{1}{4}\left(\frac{1}{2} m \omega^{2} A^{2}\right)$
$U =\frac{E}{4}$