Simple Harmonic Motion Previous Year Questions — JEE Main Physics
19 past-year Simple Harmonic Motion questions from JEE Main (Physics) — page 4.
Q31 (2026)
A simple pendulum of string length 30 cm performs 20 oscillations in 10s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ____ cm. [Assume that the mass of the pendulum remains same.]
- $120$
- $0.75$
- $7.5$
- $15$
### Related Formula
$$T = 2\pi \sqrt{\frac{l}{g}}$$
$$T \propto \sqrt{l}$$
### Core Logic
For a simple pendulum, the period is proportional to the square root of the length. If the frequency is doubled, the period is halved, which requires the length to become one-fourth of its original value.
### Step 1: Find Old and New Period
Initial time period $T_{1} = \frac{10}{20} = 0.5 \text{ s}$
New time period $T_{2} = \frac{10}{40} = 0.25 \text{ s}$
So, $T_{2} = \frac{T_{1}}{2}$
### Step 2: Calculate New Length
$$\frac{T_{1}}{T_{2}} = \sqrt{\frac{l_{1}}{l_{2}}}$$
$$2 = \sqrt{\frac{30}{l_{2}}}$$
$$4 = \frac{30}{l_{2}}$$
$$l_{2} = \frac{30}{4} = 7.5 \text{ cm}$$
### Pattern Recognition
Sees: "Oscillations doubled in same time" → Frequency is doubled → Time period is halved. Since $T \propto \sqrt{l}$, $l$ must be reduced to $\frac{1}{4}$ of the original.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q28 (2026)
A cylindrical block of mass M and area of cross section A is floating in a liquid of density $\rho$ and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is ____.
- $2\pi\sqrt{\frac{M}{\rho Ag}}$
- $\pi \sqrt{\frac{2\mathrm{M}}{\rho\mathrm{Ag}}}$
- $\pi \sqrt{\frac{\rho\mathrm{A}}{\mathrm{Mg}}}$
- $2\pi \sqrt{\frac{\rho A}{Mg}}$
### Related Formula
$$T = 2\pi \sqrt{\frac{M}{k_{\text{effective}}}}$$
$$F_{\text{buoyancy}} = \rho V g$$
### Core Logic
{{SOL_IMG1}}
At equilibrium, the buoyant force balances the weight:
$\rho Ahg = Mg$
After displacing the block downward by a small distance $x$, the net restoring force is:
$$Ma = -\rho A(h+x)g + Mg$$
$$Ma = -\rho Ahg - \rho Axg + Mg$$
Since $\rho Ahg = Mg$, this simplifies to:
$Ma = -\rho Axg$
$$a = \left(\frac{-\rho Ag}{M}\right) x$$
### Step 1: Compare with Standard SHM Equation
Comparing with $a = -\omega^2 x$, we get:
$$\omega = \sqrt{\frac{\rho Ag}{M}}$$
The time period $T$ is:
$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{M}{\rho Ag}}$$
### Pattern Recognition
For a floating body of uniform cross-section $A$, the restoring force constant is simply $k = \rho A g$. The time period is immediately $T = 2\pi \sqrt{m/k}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Class 11 Physics: Mechanical Properties of Fluids
Q46 (2026)
The displacement of a particle, executing simple harmonic motion with time period T, is expressed as $x(t) = A \sin \omega t$ , where A is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T/2\beta$ . The value of $\beta$ is ____.
### Related Formula
$$PE = \frac{1}{2} k x^2$$
$PE_{\text{max}}$ occurs when $x = \pm A$.
### Core Logic
Potential energy is maximum at the extreme positions. From $x(t) = A \sin(\omega t)$, the particle starts at the mean position ($x=0$ at $t=0$) and reaches the extreme position ($x=A$) for the first time at $t = T/4$.
### Step 1: Finding beta
Given time for maximum PE is $t = \frac{T}{4}$.
We are given $t = \frac{T}{2\beta}$.
Therefore, $\frac{T}{4} = \frac{T}{2\beta} \Rightarrow 2\beta = 4 \Rightarrow \beta = 2$.
### Pattern Recognition
Sine function SHM starts at mean, hits extreme at $T/4$, back to mean at $T/2$, negative extreme at $3T/4$, back to mean at $T$. PE is max at $T/4$ and $3T/4$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q42 (2026)
As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses $1 \text{ kg}$ and $0.2 \text{ kg}$ with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is :
{{IMG1}}
- $30$
- $27$
- $20$
- $5$
### Related Formula
$$\omega = \sqrt{\frac{k}{\mu}}$$
where $\mu$ is the reduced mass of the system given by:
$$\mu = \frac{m_1 m_2}{m_1 + m_2}$$
### Step 1: Calculate Reduced Mass
Given $m_1 = 1 \text{ kg}$ and $m_2 = 0.2 \text{ kg}$.
$$\mu = \frac{1 \times 0.2}{1 + 0.2} = \frac{0.2}{1.2}$$
$$\mu = \frac{1}{6} \text{ kg}$$
### Step 2: Calculate Angular Frequency
Given spring constant $k = 150 \text{ N/m}$.
$$\omega = \sqrt{\frac{150}{1/6}} = \sqrt{150 \times 6} = \sqrt{900}$$
$$\omega = 30 \text{ rad/s}$$
### Pattern Recognition
For any two-body spring system free of external forces, always substitute the individual mass $m$ with the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$ in the standard oscillator formula $\omega = \sqrt{k/m}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
