When the potential energy of a particle executing SHM 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 SHM 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)  (2a/3)

Tardigrade Admin · Accepted Answer

Maximum potential energy of particle executing SHM U max =(1/2) m ω2 a2 Potential energy U =(1/4) U max (given) (1/2) m ω2 y2=(1/4)((1/2) m ω2 a2) y2=(a2/4) ⇒ y=(a/2)

Q. When the potential energy of a particle executing SHM 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

$\frac{a}{4}$

$\frac{a}{3}$

$\frac{a}{2}$

$\frac{2 a}{3}$

Maximum potential energy of particle executing SHM
$U_{m a x} = \frac{1}{2} m ω^{2} a^{2}$
Potential energy $U = \frac{1}{4} U_{m a x}$ (given)
$\frac{1}{2} m ω^{2} y^{2} = \frac{1}{4} (\frac{1}{2} m ω^{2} a^{2})$
$y^{2} = \frac{a ^{2}}{4}$
$\Rightarrow y = \frac{a}{2}$