Oscillations & Waves Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Oscillations and Waves with concise, step-by-step solutions covering SHM, spring-mass systems, wave equations and periodicity.
A curated set of JEE Main 2026 Oscillations and Waves problems with clean, worked solutions to help you master simple harmonic motion, spring systems, and progressive waves.
Solutions are AI-generated and pending review.
Solution
For SHM the velocity satisfies $v^2 = \omega^2\left(A^2 - x^2\right)$.
Comparing with the given $v^2 = 50 - x^2$:
$$\omega^2 = 1 \implies \omega = 1\ \text{rad/s}$$Time period:
$$T = \frac{2\pi}{\omega} = 2\pi\ \text{s}$$Given $T = \dfrac{x}{7}$:
$$\frac{x}{7} = 2\pi \implies x = 14\pi \approx 43.98 \approx 44$$Answer: 44
Solution
Start the motion from the extreme position, so use $x = A\cos(\omega t)$ with $x = A$ at $t = 0$.
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{5}\ \text{rad/s}$$At $x = \dfrac{A}{\sqrt{2}}$:
$$A\cos(\omega t) = \frac{A}{\sqrt{2}} \implies \cos(\omega t) = \frac{1}{\sqrt{2}} \implies \omega t = \frac{\pi}{4}$$Therefore:
$$t = \frac{\pi/4}{\omega} = \frac{\pi/4}{2\pi/5} = \frac{5}{8}\ \text{s}$$Answer: $\dfrac{5}{8}$
Solution
Write the equation as $y = 5\cos(\omega t - kx)$ with:
$$\omega = 200\pi\ \text{rad/s}, \qquad k = \frac{\pi}{150}\ \text{cm}^{-1}$$Wave velocity:
$$v = \frac{\omega}{k} = \frac{200\pi}{\pi/150} = 200 \times 150 = 30000\ \text{cm/s}$$Converting to SI units:
$$v = 30000\ \text{cm/s} = 300\ \text{m/s}$$Answer: 300
Solution
Here $\omega = 50$ rad/s and phase $\phi = 50t + \dfrac{\pi}{3}$.
Particle at rest ($v = 0$): velocity $v = 50a\cos\phi$, so the first instant of rest is at the extreme, $\cos\phi = 0$:
$$50t_1 + \frac{\pi}{3} = \frac{\pi}{2} \implies 50t_1 = \frac{\pi}{6} \implies t_1 = \frac{\pi}{300}\ \text{s}$$Zero acceleration ($a = 0$): acceleration $\propto -\sin\phi$, so it vanishes at the mean position, $\sin\phi = 0$:
$$50t_2 + \frac{\pi}{3} = \pi \implies 50t_2 = \frac{2\pi}{3} \implies t_2 = \frac{2\pi}{150} = \frac{\pi}{75}\ \text{s}$$Answer: $\dfrac{\pi}{300}\,\text{s},\ \dfrac{\pi}{75}\,\text{s}$
Solution
Spring constant from the static stretch $x_0 = 2\ \text{mm} = 2\times10^{-3}$ m, $m = 0.2$ kg:
$$k = \frac{mg}{x_0} = \frac{0.2 \times 10}{2\times10^{-3}} = 1000\ \text{N/m}$$Frequency:
$$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{1000}{0.2}} = \frac{1}{2\pi}\sqrt{5000} = \frac{50\sqrt{2}}{2\pi} = \frac{25\sqrt{2}}{\pi} = \frac{5\sqrt{50}}{\pi}\ \text{Hz}$$Maximum energy in the spring occurs at maximum extension, which is the static stretch plus the amplitude:
$$x_{\max} = 2\ \text{mm} + 2\ \text{mm} = 4\ \text{mm} = 4\times10^{-3}\ \text{m}$$$$E_{\max} = \frac{1}{2}k\,x_{\max}^2 = \frac{1}{2}(1000)\left(4\times10^{-3}\right)^2 = 500 \times 16\times10^{-6} = 8\times10^{-3}\ \text{J}$$Answer: $\dfrac{5\sqrt{50}}{\pi}$ and $8\times10^{-3}$
Solution
The distance between two successive crests equals one wavelength.
From the wave equation, the wave number is:
$$k = 0.018\ \text{cm}^{-1}$$Wavelength:
$$\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{0.018} = \frac{6.28}{0.018} = 348.9\ \text{cm}$$Nearest integer:
$$\lambda \approx 349\ \text{cm}$$Answer: 349
Solution
A. $\sin^2\omega t$: Using $\sin^2\omega t = \dfrac{1 - \cos 2\omega t}{2}$, this is a single cosine of angular frequency $2\omega$, so period $T = \dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}$ and it is SHM (about the mean value $\tfrac12$). $\to$ III
B. $\sin^3(2\omega t)$: Using $\sin^3\theta = \dfrac{3\sin\theta - \sin 3\theta}{4}$ with $\theta = 2\omega t$, the lowest frequency term has angular frequency $2\omega$, giving period $\dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}$. It is a mix of two frequencies, so not SHM. $\to$ I
C. $\sin(\omega t) + \cos(\pi\omega t)$: Individual periods are $\dfrac{2\pi}{\omega}$ and $\dfrac{2\pi}{\pi\omega} = \dfrac{2}{\omega}$. Their ratio is $\dfrac{2\pi/\omega}{2/\omega} = \pi$, an irrational number, so no common period exists $\Rightarrow$ non-periodic. $\to$ IV
D. $\cos\omega t + \cos 2\omega t$: Superposition of two commensurate frequencies with fundamental period $\dfrac{2\pi}{\omega}$; a sum of distinct frequencies is not SHM. $\to$ II
Thus A-III, B-I, C-IV, D-II.
Answer: A-III, B-I, C-IV, D-II
Solution
The spring constant is inversely proportional to the natural length: $k \propto \dfrac{1}{L}$.
Cutting the spring to half its length doubles the spring constant:
$$k_2 = 2k_1$$Frequency of a spring-mass system:
$$v = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \implies v \propto \sqrt{k}$$Therefore:
$$\frac{v_2}{v_1} = \sqrt{\frac{k_2}{k_1}} = \sqrt{2}$$Answer: $\sqrt{2}$