Question

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

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

Answer 1

The correct option is D $a/2$

Given that,

Displacement $=x$

Amplitude $=a$

Potential energy $P.E=\frac{1}{2}m{\omega }^2{x}^2$

Mechanical energy $M.E=\frac{1}{2}m{\omega }^2{a}^2$

Now, the potential energy of a particles executing simple harmonic is one fourth of its maximum value during the oscillation

So,

$P.E=\frac{1}{4}M.E$

$\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{4}\times \frac{1}{2}m{\omega }^2{a}^2$

$x^2=\frac{1}{4}{a}^2$

$x=\frac{a}{2}$

Hence, the displacement of the particle from the equilibrium position in terms of its amplitude is $\frac{a}{2}$