JEE Main 2026 — Physics Waves & Oscillations

For A: $y = \sin^2 \omega t = \dfrac{1 - \cos 2\omega t}{2} = \dfrac{1}{2} - \dfrac{1}{2} \cos 2\omega t$y=sin2ωt=21−cos2ωt​=21​−21​cos2ωt

This represents simple harmonic motion (SHM) about the mean position $y = \dfrac{1}{2}$y=21​ with an angular frequency of $2\omega$2ω. The time period is $T = \dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}$T=2ω2π​=ωπ​. Thus, A matches III.

For B: $y = \sin^3(2\omega t) = \dfrac{3\sin 2\omega t - \sin 6\omega t}{4}$y=sin3(2ωt)=43sin2ωt−sin6ωt​

This is a superposition of two SHMs with angular frequencies $2\omega$2ω and $6\omega$6ω. It is periodic but not SHM. The time period is the LCM of $T_1 = \dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}$T1​=2ω2π​=ωπ​ and $T_2 = \dfrac{2\pi}{6\omega} = \dfrac{\pi}{3\omega}$T2​=6ω2π​=3ωπ​, which is $T = \dfrac{\pi}{\omega}$T=ωπ​. Thus, B matches I.

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

The angular frequencies are $\omega_1 = \omega$ω1​=ω and $\omega_2 = \pi \omega$ω2​=πω. The ratio $\dfrac{\omega_1}{\omega_2} = \dfrac{1}{\pi}$ω2​ω1​​=π1​ is an irrational number. Therefore, the function is non-periodic. Thus, C matches IV.

For D: $y = \cos \omega t + \cos 2\omega t$y=cosωt+cos2ωt

This is a superposition of two SHMs with angular frequencies $\omega$ω and $2\omega$2ω. It is periodic but not SHM. The time period is the LCM of $T_1 = \dfrac{2\pi}{\omega}$T1​=ω2π​ and $T_2 = \dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}$T2​=2ω2π​=ωπ​, which is $T = \dfrac{2\pi}{\omega}$T=ω2π​. 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