Simple Harmonic Motion Previous Year Questions — JEE Main Physics
19 past-year Simple Harmonic Motion questions from JEE Main (Physics) — page 3.
Q58 (2024)
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4 \mathrm{~m}$, then the time period of small oscillations will be ________ $\mathrm{s}$. [take $\mathrm{g} = \pi^2 \mathrm{~m/s}^2$]
### Related Formula
$$g' = \frac{GM}{(R+h)^2} = g\left(\frac{R}{R+h}\right)^2$$
$$T = 2\pi \sqrt{\frac{l}{g'}}$$
### Core Logic
The distance of the pendulum from the earth's surface is $h = R$.
The acceleration due to gravity at this height $g'$ is:
$$g' = g\left(\frac{R}{R+R}\right)^2 = g\left(\frac{1}{2}\right)^2 = \frac{g}{4}$$
### Step 1: Calculate Time Period
The time period of the simple pendulum is:
$$T = 2\pi \sqrt{\frac{l}{g'}}$$
Substitute $l = 4 \mathrm{~m}$ and $g' = \frac{g}{4}$:
$$T = 2\pi \sqrt{\frac{4}{g/4}} = 2\pi \sqrt{\frac{16}{g}}$$
Given $g = \pi^2$:
$$T = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi \left(\frac{4}{\pi}\right) = 8 \mathrm{~s}$$
### Pattern Recognition
At $h=R$, $g$ drops to $g/4$. Thus, the time period doubles compared to its surface value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Class 11 Physics: Gravitation
Q59 (2024)
The time period of simple harmonic motion of mass M in the given figure is $\pi \sqrt{\frac{\alpha M}{5K}}$, where the value of $\alpha$ is
{{IMG1}}
### Related Formula
Springs in parallel:
$k_p = k_1 + k_2$
Springs in series:
$$\frac{1}{k_s} = \frac{1}{k_p} + \frac{1}{k_3}$$
Time period:
$$T = 2\pi \sqrt{\frac{M}{k_{eq}}}$$
### Core Logic
Evaluate the equivalent spring constant of the given setup by first resolving the parallel configuration, then the series configuration.
### Step 1: Equivalent Spring Constant
Based on the standard layout of the problem (as implied by the solution text), let's assume one section is $2k$ and $k$ in parallel... Wait, the solution implies $\frac{2k \cdot k}{3k} + k$, which represents a series combination of $2k$ and $k$, followed by a parallel combination with another $k$. Let's trust the official PDF's algebraic step: $k_{eq} = \frac{2k \cdot k}{3k} + k$.
This evaluates to:
$$k_{eq} = \frac{2k}{3} + k = \frac{5k}{3}$$
### Step 2: Time Period Calculation
$$T = 2\pi \sqrt{\frac{M}{k_{eq}}}$$
$$T = 2\pi \sqrt{\frac{M}{5k/3}} = 2\pi \sqrt{\frac{3M}{5k}}$$
### Step 3: Matching the Format
Bring the 2 inside the square root as 4:
$$T = \pi \sqrt{\frac{4 \times 3M}{5k}} = \pi \sqrt{\frac{12M}{5k}}$$
Comparing with $\pi \sqrt{\frac{\alpha M}{5k}}$:
$\alpha = 12$
### Pattern Recognition
Whenever an external coefficient like 2 (from $2\pi$) is pushed inside the square root to match a format, it squares. Don't forget $2 \to 4$, turning $3M$ into $12M$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q59 (2024)
A particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is ______.
### Related Formula
$$v = \omega \sqrt{A^2 - x^2}$$
### Core Logic
At displacement $x = \frac{2A}{3}$, the initial velocity is:
$$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2}$$
$$v = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}}$$
$$v = \frac{\sqrt{5}A\omega}{3}$$
### Step 2: Velocity Tripled
The speed is now increased to $v' = 3v$ at the same position $x = \frac{2A}{3}$.
$$v' = 3 \left(\frac{\sqrt{5}A\omega}{3}\right) = \sqrt{5}A\omega$$
This new velocity corresponds to a new amplitude $A'$:
$$v' = \omega \sqrt{(A')^2 - x^2}$$
$$\sqrt{5}A\omega = \omega \sqrt{(A')^2 - \left(\frac{2A}{3}\right)^2}$$
### Step 3: Finding new Amplitude
Squaring both sides:
$$5A^2 = (A')^2 - \frac{4A^2}{9}$$
$$(A')^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9}$$
$$(A')^2 = \frac{49A^2}{9}$$
$$A' = \frac{7A}{3}$$
Comparing with $\frac{nA}{3}$, we get $n = 7$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q33 (2026)
The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of $176 \text{ rad/s}$. The frequency of this simple harmonic oscillator is ________ Hz. $[\text{take } \pi = \frac{22}{7}]$
- $14$
- $88$
- $28$
- $176$
### Related Formula
$$\omega_k = 2\pi f_k$$
$$f_{\text{oscillator}} = \frac{f_k}{2}$$
### Core Logic
For a simple harmonic oscillator with an angular frequency $\omega$, the kinetic energy oscillates at double the angular frequency ($2\omega$).
The given angular frequency of kinetic energy oscillation is $\omega_k = 176 \text{ rad/s}$.
### Step 1: Finding Kinetic Energy Frequency
First, find the frequency of kinetic energy oscillations ($f_k$):
$$f_k = \frac{\omega_k}{2\pi} = \frac{176}{2 \times \frac{22}{7}}$$
$$f_k = \frac{176 \times 7}{44} = 4 \times 7 = 28 \text{ Hz}$$
### Step 2: Final Conclusion
Since kinetic energy oscillates at twice the frequency of the oscillator itself:
$$f_{\text{oscillator}} = \frac{f_k}{2} = \frac{28}{2} = 14 \text{ Hz}$$
### Pattern Recognition
KE and PE always oscillate at twice the frequency ($2\nu$) of the underlying SHM ($v$) due to the $\sin^2(\omega t)$ or $\cos^2(\omega t)$ terms resolving to $(1 - \cos(2\omega t))/2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q27 (2026)
Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T?
- (1)
- (2)
- (3)
- (4)
### Related Formula
$$T = 2\pi \sqrt{\frac{L}{g}}$$
### Core Logic
Squaring both sides of the simple pendulum equation:
$$T^2 = \frac{4\pi^2 L}{g}$$
Inverting $T^2$ to express $1/T^2$ as a function of length $L$:
$$\frac{1}{T^2} = \frac{g}{4\pi^2 L}$$
This demonstrates that $\frac{1}{T^2} \propto \frac{1}{L}$. The relationship between $1/T^2$ and $L$ represents a rectangular hyperbola curve decreasing asymptotically.
{{SOL_IMG1}}
### Step 1: Final Conclusion
Plot (2) correctly depicts the rectangular hyperbola curve for $\frac{1}{T^2}$ versus $L$.
### Pattern Recognition
Sees: Graph of $1/T^2$ vs $L$.
Formula check: $T \propto \sqrt{L} \implies T^2 \propto L \implies 1/T^2 \propto 1/L$, which is inversely proportional (hyperbolic curve).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Simple Harmonic Motion
