Oscillations (SHM): JEE Main 2026 PYQs

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Simple Harmonic Motion - velocity and acceleration conditions

Answer: 7

Key idea: Simple Harmonic Motion - Velocity-Displacement Relation

1. Given velocity equation: $v^2 = 50 - x^2$, which can be written as $v^2 + x^2 = 50$
2. For SHM, velocity is given by $v^2 = \omega^2(A^2 - x^2)$ where $A$ is amplitude
3. Comparing with given equation: $\omega^2(A^2 - x^2) = 50 - x^2$
4. This gives $\omega^2 = 1$ (coefficient of $x^2$) and $\omega^2 A^2 = 50$ (constant term)
5. Therefore $A^2 = 50$, so amplitude $A = \sqrt{50} = 5\sqrt{2} \approx 7.07$
6. The question asks for the value of $x$, which refers to the amplitude
7. Answer: $x = 7$ (rounded to nearest integer)

Answer: Option (3)

Key idea: Physical Pendulum - Time Period using Parallel Axis Theorem

1. Identify this as a physical pendulum with axis at point A on the edge, distance from pivot to center of mass: $d = R$
2. Find moment of inertia about center: $I_{center} = \frac{1}{2}MR^2$
3. Apply parallel axis theorem: $I_A = I_{center} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$
4. Use physical pendulum formula: $T = 2\pi\sqrt{\frac{I_A}{Mgd}}$
5. Substitute values: $T = 2\pi\sqrt{\frac{\frac{3}{2}MR^2}{MgR}} = 2\pi\sqrt{\frac{3R}{2g}}$

Answer: Option (1)

Key idea: SHM of a dipole in electric field; frequency depends on square root of field magnitude

1. For a dipole in electric field, the restoring torque for small angle $\theta$ is $\tau = -pE\theta$, giving angular frequency $\omega = \sqrt{\frac{pE}{I}}$
2. Initial field: $\vec{E}_1 = E_0\hat{x}$, so $E_1 = E_0$ and $f_1 \propto \sqrt{E_0}$
3. Total field after adding $\vec{E}_2$: $\vec{E}_{total} = E_0\hat{x} + 2E_0\hat{y} + 2E_0\hat{z}$
4. Magnitude: $E_{total} = \sqrt{E_0^2 + 4E_0^2 + 4E_0^2} = \sqrt{9E_0^2} = 3E_0$
5. Frequency ratio: $\frac{f_2}{f_1} = \sqrt{\frac{3E_0}{E_0}} = \sqrt{3}$
6. Percentage change: $(\sqrt{3} - 1) \times 100\% = (1.732 - 1) \times 100\% = 73\%$

Answer: Option (1)

Key idea: Periodic Motion and Simple Harmonic Motion (SHM) - Identifying periodicity and SHM characteristics using trigonometric identities

1. For **A**: Use $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$. Period is $T = \frac{\pi}{\omega}$. About mean position $\frac{1}{2}$, it satisfies SHM equation $\frac{d^2y}{dt^2} = -4\omega^2 y$ → **III**
2. For **B**: Use $\sin^3(2\omega t) = \frac{3\sin(2\omega t) - \sin(6\omega t)}{4}$. Period is LCM of $\frac{\pi}{\omega}$ and $\frac{\pi}{3\omega}$ which is $\frac{\pi}{\omega}$. Sum of different frequencies → NOT SHM → **I**
3. For **C**: $\sin(\omega t)$ has period $\frac{2\pi}{\omega}$ and $\cos(\pi\omega t)$ has period $\frac{2}{\omega}$. Ratio of periods is $\pi$ (irrational) → **Non-periodic** → **IV**
4. For **D**: $\cos(\omega t)$ has period $\frac{2\pi}{\omega}$ and $\cos(2\omega t)$ has period $\frac{\pi}{\omega}$. LCM gives $T = \frac{2\pi}{\omega}$. Sum of different frequencies → NOT SHM → **II**
5. Match: A-III, B-I, C-IV, D-II

Answer: 0.625

Key idea: Simple Harmonic Motion - Time calculation between two positions

1. Write the SHM equation: $x(t) = A\cos(\omega t)$ where $\omega = \frac{2\pi}{T} = \frac{2\pi}{5}$ rad/s
2. At $t = 0$, particle is at $x = A$ (extreme position), so phase constant $\phi = 0$
3. For position $x = \frac{A}{\sqrt{2}}$: Set $\frac{A}{\sqrt{2}} = A\cos\left(\frac{2\pi t}{5}\right)$
4. Solve: $\cos\left(\frac{2\pi t}{5}\right) = \frac{1}{\sqrt{2}}$ gives $\frac{2\pi t}{5} = \frac{\pi}{4}$
5. Calculate time: $t = \frac{5\pi}{8\pi} = \frac{5}{8} = 0.625$ sec

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Answer: Option (3)

Key idea: Spring constant inversely proportional to length; frequency of mass-spring system

1. Write the frequency formula: $v = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$
2. Recognize that spring constant is inversely proportional to length: $k \propto \frac{1}{L}$
3. When spring is cut to half length, new spring constant: $k' = 2k$
4. Calculate frequency ratio: $\frac{v_2}{v_1} = \sqrt{\frac{k'}{k}} = \sqrt{\frac{2k}{k}}$
5. Simplify to get: $\frac{v_2}{v_1} = \sqrt{2}$