JEE Main 2026 — Physics Waves & Oscillations

In a simple harmonic oscillator, if the displacement is $x = A \sin(\omega t + \phi)$x=Asin(ωt+ϕ), the velocity is $v = A\omega \cos(\omega t + \phi)$v=Aωcos(ωt+ϕ).

The kinetic energy is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t + \phi)$K=21​mv2=21​mA2ω2cos2(ωt+ϕ).

Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$cos2θ=21+cos2θ​, we get $K = \frac{1}{4}m A^2 \omega^2 [1 + \cos(2\omega t + 2\phi)]$K=41​mA2ω2[1+cos(2ωt+2ϕ)].

The angular frequency of the kinetic energy oscillation is $\omega_{KE} = 2\omega$ωKE​=2ω, where $\omega$ω is the angular frequency of the SHM.

Given $\omega_{KE} = 176$ωKE​=176 rad/s, we have $2\omega = 176$2ω=176, which gives $\omega = 88$ω=88 rad/s.

The frequency of the simple harmonic oscillator is $f = \frac{\omega}{2\pi}$f=2πω​.

Substituting the values: $f = \frac{88}{2 \times \frac{22}{7}} = \frac{88 \times 7}{44} = 2 \times 7 = 14$f=2×722​88​=4488×7​=2×7=14 Hz.