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

Question

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is (A) (a/4) (B) (a/3) (C) (a/2) (D) (2 a/3)

Tardigrade Admin · Accepted Answer

As, (U/U max ) =((1/2) m ω2 y2/(1)2 m ω2 a2)=(1/4) (y2/a2) =(1/4) ⇒ y =(a/2)

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

$\frac{a}{4}$

$\frac{a}{3}$

$\frac{a}{2}$

$\frac{2 a}{3}$

As, $\frac{U}{U _{m a x}} = \frac{\frac{1}{2} m ω ^{2} y ^{2}}{\frac{1}{2} m ω ^{2} a ^{2}} = \frac{1}{4}$
$\frac{y ^{2}}{a ^{2}} = \frac{1}{4}$
$\Rightarrow y = \frac{a}{2}$

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

4a​

3a​

2a​

32a​