JEE Main 2026 — Oscillations Question with Solution

For A: $y={\sin }^2\omega t=\frac{1-\cos 2\omega t}{2}=\frac{1}{2}-\frac{1}{2}\cos 2\omega t$

This represents simple harmonic motion (SHM) about the mean position $y=\frac{1}{2}$ with an angular frequency of $2\omega$. The time period is $T=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$. Thus, A matches III.

For B: $y={\sin }^3(2\omega t)=\frac{3\sin 2\omega t-\sin 6\omega t}{4}$

This is a superposition of two SHMs with angular frequencies $2\omega$ and $6\omega$. It is periodic but not SHM. The time period is the LCM of $T_1=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$ and $T_2=\frac{2\pi }{6\omega }=\frac{\pi }{3\omega }$, which is $T=\frac{\pi }{\omega }$. Thus, B matches I.

For C: $y=\sin (\omega t)+\cos (\pi \omega t)$

The angular frequencies are ${\omega }_1=\omega$ and ${\omega }_2=\pi \omega$. The ratio $\frac{{\omega }_1}{{\omega }_2}=\frac{1}{\pi }$ is an irrational number. Therefore, the function is non-periodic. Thus, C matches IV.

For D: $y=\cos \omega t+\cos 2\omega t$

This is a superposition of two SHMs with angular frequencies $\omega$ and $2\omega$. It is periodic but not SHM. The time period is the LCM of $T_1=\frac{2\pi }{\omega }$ and $T_2=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$, which is $T=\frac{2\pi }{\omega }$. Thus, D matches II.

The correct matching is A-III, B-I, C-IV, D-II.

Answer: A-III, B-I, C-IV, D-II