Question

When the potential energy of a particle executing simple harmonic motion is one-fourth of the maximum value during the oscillation, its displacement from the equilibrium position in terms of amplitude ‘a’ is

• A. $a/4$
• B. $a/3$
• C. $a/2$
• D. $2a/3$

Answer 1

Answer: (C)

$a/2$

Answer 2

According to problem potential energy =$\frac{1}{4}$maximum Energy

⇒$\frac{1}{2}m\omega^{2}y^{2} = \frac{1}{4}\left( \frac{1}{2}m\omega^{2}a^{2} \right)$⇒$y^{2} = \frac{a^{2}}{4}$⇒$y = a/2$