Simple Harmonic Motion Previous Year Questions — JEE Main Physics
19 past-year Simple Harmonic Motion questions from JEE Main (Physics) — page 2.
Q12 (2025)
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Knowing initial position $x_0$ and initial momentum $p_0$ is enough to determine the position and momentum at any time $t$ for a simple harmonic motion with a given angular frequency $\omega$ .
Reason (R) : The amplitude and phase can be expressed in terms of $x_0$ and $p_0$ .
In the light of the above statements, choose the correct answer from the options given below:
- Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
- (A) is false but (R) is true.
- (A) is true but (R) is false.
- Both (A) and (R) are true and (R) is the correct explanation of (A).
### Related Formula
The general solution tracking position in SHM is given by :
$$x(t) = A \sin(\omega t + \phi)$$
Momentum formula:
$$p(t) = m v(t) = m A \omega \cos(\omega t + \phi)$$
### Core Logic
At $t = 0$ :
1) $x_0 = A \sin \phi$
2) $p_0 = m A \omega \cos \phi$
Dividing equation (1) by (2) reveals the phase \angle relationship :
$$\tan \phi = \left(\frac{x_0}{p_0}\right) m \omega \implies \phi = \tan^{-1}\left(\frac{m \omega x_0}{p_0}\right)$$
Squaring and combining both equations isolates amplitude $A$:
$$A = \frac{\sqrt{(m \omega x_0)^2 + p_0^2}}{m \omega}$$
Since both amplitude $A$ and phase constant $\phi$ are explicitly fixed by the initial conditions $x_0$ and $p_0$, the kinematic layout of the state space is completely specified for any future time parameter $t$. This validates that Reason (R) perfectly explains Assertion (A).
### Pattern Recognition
SHM is a second-order differential equation. Any system of second-order equations requires exactly two independent boundary conditions (e.g., initial position and velocity/momentum) to completely specify unique path tracks.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q1280 (2025)
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.
Reason (R) : Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.
In the light of the above statements, choose the most appropriate answer from the options given below:
- $\text{Both (A) and (R) are true but (R) is not the correct explanation of (A).}$
- $\text{Both (A) and (R) are true and (R) is the correct explanation of (A).}$
- $\text{(A) is true but (R) is false.}$
- $\text{(A) is false but (R) is true.}$
### Related Formula
$$T = 2\pi\sqrt{\frac{l}{g}}$$
### Core Logic
As altitude $h$ increases at the top of a mountain, acceleration due to gravity $g$ drops down according to :
$$g = \frac{g_0 R^2}{(R+h)^2}$$
Since $T \propto \frac{1}{\sqrt{g}}$, a decreased $g$ directly makes the time period $T$ longer
### Pattern Recognition
Higher altitude $\implies$ smaller gravity field $\implies$ slower pendulum oscillations $\implies$ longer period
### Chapter Mix
Class 11 Physics: Oscillations
Class 11 Physics: Gravitation
Q1465 (2024)
A simple harmonic oscillator has an amplitude $A$ and time period $6\pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x = A$ to $x = \frac{\sqrt{3}}{2}A$ will be $\frac{\pi}{x}$ s, where $x =$ ________.
### Related Formula
For a simple harmonic oscillator starting from the mean position at $t = 0$:
$$x(t) = A \sin(\omega t)$$
where $\omega = \frac{2\pi}{T}$ is the angular frequency.
### Core Logic
Given:
* Time period, $T = 6\pi\text{ s} \implies \omega = \frac{2\pi}{6\pi} = \frac{1}{3}\text{ rad/s}$
We need to find the time taken to travel from $x = A$ to $x = \frac{\sqrt{3}}{2}A$.
Let's calculate the times from the mean position ($x=0$) to both positions:
1. Time $t_1$ to reach $x = A$:
$$A = A \sin(\omega t_1) \implies \sin(\omega t_1) = 1 \implies \omega t_1 = \frac{\pi}{2}$$
2. Time $t_2$ to reach $x = \frac{\sqrt{3}}{2}A$:
$$\frac{\sqrt{3}}{2}A = A \sin(\omega t_2) \implies \sin(\omega t_2) = \frac{\sqrt{3}}{2} \implies \omega t_2 = \frac{\pi}{3}$$
### Step 1: Calculate the Time Difference
The time required to travel between these two points is:
$$\Delta t = t_1 - t_2 = \frac{\pi/2}{\omega} - \frac{\pi/3}{\omega} = \frac{\pi}{6\omega}$$
Substitute $\omega = \frac{1}{3}$:
$$\Delta t = \frac{\pi}{6 \times (1/3)} = \frac{\pi}{2}\text{ seconds}$$
Comparing this to $\frac{\pi}{x}\text{ s}$, we find:
$x = 2$
{{SOL_IMG1}}
### Pattern Recognition
Alternatively, using a phasor diagram, the angle swept in moving from $A$ to $\frac{\sqrt{3}}{2}A$ is $\phi = \frac{\pi}{6}$. Therefore, $\Delta t = \frac{\phi}{\omega} = \frac{\pi/6}{1/3} = \frac{\pi}{2}$ s.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q57 (2024)
A particle executes simple harmonic motion with an amplitude of $4\text{ cm}$. At the mean position, the velocity of the particle is $10\text{ cm/s}$. The distance of the particle from the mean position when its speed becomes $5\text{ cm/s}$ is $\sqrt{\alpha}\text{ cm}$, where $\alpha = $ ______.
### Related Formula
$$v_{\text{max}} = A\omega$$
$$v = \omega \sqrt{A^2 - x^2}$$
### Core Logic
From the mean position maximum criteria ($v_{\text{max}} = 10\text{ cm/s}$, $A = 4\text{ cm}$):
$$10 = 4\omega \implies \omega = \frac{10}{4} = \frac{5}{2}\text{ rad/s}$$
### Step 1: Substitute parameters into speed equation
Set the target speed $v = 5\text{ cm/s}$:
$$5 = \frac{5}{2} \sqrt{4^2 - x^2}$$
$$2 = \sqrt{16 - x^2}$$
Squaring both sides:
$$4 = 16 - x^2 \implies x^2 = 12 \implies x = \sqrt{12}\text{ cm}$$
### Step 2: Compare to target format
Comparing $\sqrt{12}$ to $\sqrt{\alpha}$ directly shows:
$\alpha = 12$
### Pattern Recognition
Halving the peak harmonic velocity maps spatial points directly to $\frac{\sqrt{3}}{2}A$ values through standard trigonometric projection balances.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
Q58 (2024)
When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\frac{\mathrm{x}}{8}$, where $\mathrm{x} =$ ________.
### Related Formula
Total energy ($E$) in simple harmonic motion is:
$$E = \frac{1}{2} k A^2$$
Potential energy ($U$) at displacement $y$ is:
$$U = \frac{1}{2} k y^2$$
Kinetic energy ($KE$) is the remaining energy:
$KE = E - U$
### Core Logic
Given the displacement is one-third of the amplitude:
$$y = \frac{A}{3}$$
Substituting this displacement into the potential energy expression:
$$U = \frac{1}{2} k \left(\frac{A}{3}\right)^2 = \frac{1}{9} \left( \frac{1}{2} k A^2 \right) = \frac{E}{9}$$$
### Step 1: Calculate Kinetic Energy
The kinetic energy is:
$$KE = E - \frac{E}{9} = \frac{8E}{9}$$
### Step 2: Find the Energy Ratio
The ratio of total energy to kinetic energy is:
$$\frac{E}{KE} = \frac{E}{\frac{8E}{9}} = \frac{9}{8}$$
Comparing this with the given target form $\frac{x}{8}$:
$x = 9$
Therefore, the value of $x$ is $9$.
### Pattern Recognition
Since $U \propto y^2$, if displacement scales by a fraction $\frac{1}{n}$, potential energy scales instantly by $\frac{1}{n^2}$. The remaining kinetic energy is naturally $1 - \frac{1}{n^2}$, giving an immediate shortcut for the total-to-kinetic ratio: $\frac{n^2}{n^2 - 1}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Oscillations
