JEE Main 2026 — Oscillations Question with Solution

In a simple harmonic oscillator, if the displacement is $x=A\sin (\omega t+\varphi )$, the velocity is $v=A\omega \cos (\omega t+\varphi )$.
The kinetic energy is given by $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t+\varphi )$.
Using the identity ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$, we get $K=\frac{1}{4}m{A}^2{\omega }^2[1+\cos (2\omega t+2\varphi )]$.
The angular frequency of the kinetic energy oscillation is ${\omega }_{KE}=2\omega$, where $\omega$ is the angular frequency of the SHM.
Given ${\omega }_{KE}=176$ rad/s, we have $2\omega =176$, which gives $\omega =88$ rad/s.
The frequency of the simple harmonic oscillator is $f=\frac{\omega }{2\pi }$.
Substituting the values: $f=\frac{88}{2\times \frac{22}{7}}=\frac{88\times 7}{44}=2\times 7=14$ Hz.