Simple Harmonic Motion Previous Year Questions — JEE Main Physics
19 past-year Simple Harmonic Motion questions from JEE Main (Physics).
Q84 (2025)
A particle is subjected to two simple harmonic motions as:
$$x_1 = \sqrt{7}\sin 5t\mathrm{~cm}$$
and
$$x_2 = 2\sqrt{7}\sin\left(5t + \frac{\pi}{3}\right)\mathrm{~cm}$$
where $x$ is displacement and $t$ is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2}\mathrm{~ms}^{-2}$. The value of $x$ is:
- 175
- $25\sqrt{7}$
- $5\sqrt{7}$
- 125
### Related Formula
$$A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos\phi}$$
$$a_{\max} = \omega^2 A$$
### Core Logic
Both SHMs share the same frequency $\omega = 5\mathrm{~rad/s}$ with phase difference $\phi = \frac{\pi}{3}$.
Amplitudes: $A_1 = \sqrt{7}\mathrm{~cm}$, $A_2 = 2\sqrt{7}\mathrm{~cm}$.
The combined amplitude is:
$$A = \sqrt{(\sqrt{7})^2 + (2\sqrt{7})^2 + 2(\sqrt{7})(2\sqrt{7})\cos\left(\frac{\pi}{3}\right)}$$
$$A = \sqrt{7 + 28 + 2(7)(2)\left(\frac{1}{2}\right)} = \sqrt{35 + 14} = \sqrt{49} = 7\mathrm{~cm}$$
Converting to meters:
$$A = 0.07\mathrm{~m}$$
The maximum acceleration is:
$$a_{\max} = \omega^2 A = (5)^2 \times 0.07 = 25 \times 0.07 = 1.75\mathrm{~ms}^{-2}$$
Equating to $x \times 10^{-2}\mathrm{~ms}^{-2}$:
$$1.75 = x \times 10^{-2} \implies x = 175$$
### Step 1: Final Conclusion
The value of $x$ is 175.
### Pattern Recognition
When superposing two SHMs of identical frequency, use the standard phasor addition formula to obtain the combined amplitude $A$. Then compute maximum velocity $v_{\max} = \omega A$ or maximum acceleration $a_{\max} = \omega^2 A$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Simple Harmonic Motion
Q10 (2025)
Two bodies A and B of equal mass are suspended from two massless springs of spring constant $k_{1}$ and $k_{2}$ , respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is:
- $\sqrt{\frac{k_1}{k_2}}$
- $\frac{\mathrm{k}_1}{\mathrm{k}_2}$
- $\frac{\mathrm{k}_2}{\mathrm{k}_1}$
- $\sqrt{\frac{k_{2}}{k_{1}}}$
### Related Formula
$$v_{\max} = A\omega$$
$$\omega = \sqrt{\frac{k}{m}}$$
### Core Logic
The maximum velocity in vertical SHM happens at the mean equilibrium position and is given by $v_{\max} = A\omega$.
Given $m_A = m_B = m$ and $A_A = A_B = A$:
$$v_A = A \omega_1 = A \sqrt{\frac{k_1}{m}}$$
$$v_B = A \omega_2 = A \sqrt{\frac{k_2}{m}}$$
Taking the ratio:
$$\frac{v_A}{v_B} = \frac{\omega_1}{\omega_2} = \frac{\sqrt{\frac{k_1}{m}}}{\sqrt{\frac{k_2}{m}}} = \sqrt{\frac{k_1}{k_2}}$$
### Pattern Recognition
Since maximum velocity is directly proportional to angular frequency $\omega$ for equal amplitudes, and $\omega \propto \sqrt{k}$, the velocity ratio directly yields the square root of the spring constant ratio.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q917 (2025)
Two simple pendulums having lengths $l_{1}$ and $l_{2}$ with negligible string mass undergo angular displacements $\theta_{1}$ and $\theta_{2}$, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?
- $\theta_{1}l_{2}^{2} = \theta_{2}l_{1}^{2}$
- $\theta_{1}l_{1} = \theta_{2}l_{2}$
- $\theta_{1}l_{1}^{2} = \theta_{2}l_{2}^{2}$
- $\theta_{1}l_{2} = \theta_{2}l_{1}$
### Related Formula
Angular acceleration definition for a simple pendulum swinging at small angle $\theta$:
$$\alpha = -\omega^2 \theta$$
where:
$$\omega = \sqrt{\frac{g}{l}} \implies \omega^2 = \frac{g}{l}$$
Hence, the magnitude of angular acceleration is:
$$\alpha = \frac{g}{l}\theta$$
### Core Logic
Given that angular accelerations are exactly identical in magnitude ($\alpha_1 = \alpha_2$):
$$\frac{g}{l_1}\theta_1 = \frac{g}{l_2}\theta_2$$
### Step 1: Simplify Expression
Cancelling out the constant gravitational acceleration $g$:
$$\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2} \implies \theta_1 l_2 = \theta_2 l_1$$
### Pattern Recognition
Angular acceleration scales inversely with length for a fixed angular displacement ($\alpha \propto \frac{\theta}{l}$). To maintain identical angular accelerations, the ratio $\frac{\theta}{l}$ must remain constant, yielding the cross-multiplication relation $\theta_1 l_2 = \theta_2 l_1$.
### Evaluation Rubric / Model Answer
Option D: $\theta_{1}l_{2} = \theta_{2}l_{1}$
### Chapter Mix
Class 11 Physics: Oscillations
Q19 (2025)
A particle oscillates along the x-axis according to the law, $\mathrm{x}(t) = \mathrm{x}_{0} \sin^{2}\left(\frac{t}{2}
ight)$ where $x_{0} = 1 \, m$ . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph.
- Graph (1)
- Graph (2)
- Graph (3)
- Graph (4)
### Related Formula
$$x(t) = x_0 \sin^2\left(\frac{t}{2}
ight) = x_0 \left(\frac{1 - \cos t}{2}
ight)$$
### Core Logic
Given $x_0 = 1\ \mathrm{m}$, the position simplifies to:
$$x = \frac{1 - \cos t}{2} \implies 2x - 1 = -\cos t \implies \cos t = 1 - 2x$$
Differentiating $x(t)$ to find velocity $v$:
$$v = \frac{dx}{dt} = \frac{1}{2}\sin t$$
Kinetic energy $K$ is proportional to $v^2$:
$$K = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{1}{4}\sin^2 t
ight) = \frac{m}{8}(1 - \cos^2 t)$$
Substituting $\cos t = 1 - 2x$:
$$K = \frac{m}{8}\left[1 - (1 - 2x)^2
ight] = \frac{m}{8}\left[1 - (1 - 4x + 4x^2)
ight] = \frac{m}{8}(4x - 4x^2) = \frac{m}{2}(x - x^2)$$
This is an inverted parabola ($K \propto x - x^2$) passing through $x=0$ and $x=1$, reaching a peak at $x = 1/2$. This matches Graph (1).
### Pattern Recognition
The position oscillates purely within the boundary interval $[0, 1]$. Kinetic energy peaks perfectly at the equilibrium midpoint $x = 1/2$ where speed is maximum.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q13 (2025)
A particle is executing simple harmonic motion with time period 2s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then $\frac{D}{d}$ is:-
- $\frac{15}{4}$
- 25
- 10
- $\frac{16}{5}$
### Related Formula
In Simple Harmonic Motion:
* In one full oscillation period $T$, total distance = $4A$, and net displacement = $0$.
* In a quarter period $\frac{T}{4}$ starting from equilibrium, distance = $A$, and displacement = $A$.
### Core Logic
Find the total number of cycles completed in the given time :
$$\text{Cycles} = \frac{t}{T} = \frac{12.5\text{ s}}{2\text{ s}} = 6.25\text{ cycles} = 6 \text{ full cycles} + 0.25\text{ cycle}$$
### Step 1: Evaluation of Distance and Displacement
Calculate the total distance $D$ covered :
$$D = 6 \times (4A) + 1A = 24A + A = 25A$$
Given amplitude $A = 1\text{ cm}$ , we get $D = 25\text{ cm}$ .
Calculate the net displacement $d$:
The first $6$ complete cycles return the particle to equilibrium with zero displacement. The remaining $0.25$ cycle (which is a quarter period $\frac{T}{4}$) moves it from the center to its maximum amplitude value :
$$d = 1A = 1\text{ cm}$$
Taking their ratio:
$$\frac{D}{d} = \frac{25}{1} = 25$$
### Pattern Recognition
Fractional cycles modulate ratios abruptly. Always break the path index down into integer full cycles plus the final residual interval.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
