Rankers Physics

Energy in SHM: Practice Problem & Solution

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is: (2003)
$\frac{2}{3} E$
$\frac{1}{8} E$
$\frac{1}{4} E$
$\frac{1}{2} E$

Solution Explained:

To solve this problem, we apply the core principles of Energy in SHM. Understanding the underlying formula is key to arriving at the correct answer below:

Total energy $E = \frac{1}{2}kA^2$. Halfway to the endpoint means $x = A/2$.\nPotential energy $$U = \frac{1}{2}kx^2 = \frac{1}{2}k(A/2)^2 = \frac{1}{4}(\frac{1}{2}kA^2)$$.\nTherefore, $U = E/4$.

Leave a Reply

Your email address will not be published. Required fields are marked *