JEE Main SHM PYQ

1. (JEE Main 2023 (Online) 15th April Morning Shift)

In a linear Simple Harmonic Motion (SHM)

(A) Restoring force is directly proportional to the displacement.

(B) The acceleration and displacement are opposite in direction.

(C) The velocity is maximum at mean position.

(D) The acceleration is minimum at extreme points.

Choose the correct answer from the options given below:

A. (A), (B) and (D) only

B. (C) and (D) only

C. (A), (B) and (C) Only

D. (A), (C) and (D) only

Correct Answer is Option (C)

The correct options are:

(A) Restoring force is directly proportional to the displacement. - True (this is a defining characteristic of SHM)

(B) The acceleration and displacement are opposite in direction. - True (the acceleration is proportional to the displacement but in the opposite direction)

(C) The velocity is maximum at mean position. - True (the velocity is zero at the extreme positions and reaches a maximum at the mean position)

(D) The acceleration is minimum at extreme points. - False (the acceleration is maximum at the extreme points and zero at the mean position)

2. (JEE Main 2023 (Online) 10th April Morning Shift)

A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is $x = \frac{A}{2}$ and it moves along positive x-axis. The displacement of particle in time t is $x = A \sin (w t + δ)$ , then the value of $δ$ will be

A. $\frac{π}{2}$

B. $\frac{π}{3}$

C. $\frac{π}{4}$

D. $\frac{π}{6}$

Correct Answer is Option (D)

The initial condition states that the particle is at position $x = \frac{A}{2}$ at $t = 0$ .

If we substitute these initial conditions into the equation for the displacement of the particle:

$x = A \sin (w t + δ)$

We have:

$\frac{A}{2} = A \sin (δ)$

Dividing both sides by $A$ gives us:

$\frac{1}{2} = \sin (δ)$

The angle whose sine is $\frac{1}{2}$ is $δ = \frac{π}{6}$ radians (or 30 degrees).

Therefore, the correct answer is $δ = \frac{π}{6}$ .

3. (JEE Main 2023 (Online) 8th April Evening Shift)

For particle P revolving round the centre O with radius of circular path $r$ and angular velocity $ω$ , as shown in below figure, the projection of OP on the $x$ -axis at time $t$ is

JEE Main 2023 (Online) 8th April Evening Shift Physics - Simple Harmonic Motion Question 3 English

A. $x (t) = \cos (ω t - \frac{π}{6} ω)$

B. $x (t) = \cos (ω t)$

C. $x (t) = r \cos (ω t + \frac{π}{6})$

D. $x (t) = r \sin (ω t + \frac{π}{6})$

Correct Answer is Option (C)

JEE Main 2023 (Online) 8th April Evening Shift Physics - Simple Harmonic Motion Question 3 English Explanation
After time $t$ , the angular displacement will be

$θ = ω t$

Total angular displacement from $x$ -axis.

$θ_{total} = ω t + \frac{π}{6} (∵ 30^{\circ} = \frac{π}{6})$

Now, OP has two component

The horizontal component will be the projection along $x$ -axis

$= r \cos (θ_{Total}) = r \cos (ω t + \frac{π}{6})$

4. (JEE Main 2023 (Online) 25th January Evening Shift)

A particle executes simple harmonic motion between $x = - A$ and $x = + A$ . If time taken by particle to go from $x = 0$ to $\frac{A}{2}$ is 2 s; then time taken by particle in going from $x = \frac{A}{2}$ to A is

A. 4 s

B. 1.5 s

C. 3 s

D. 2 s

Correct Answer is Option (A)

$x = A \sin (ω t)$

$\begin{aligned} x = \frac{A}{2} = A \sin (ω t) \\ \frac{1}{2} = \sin (ω t) \\ t = (\frac{π}{6 ω}) = 2 \\ \frac{π}{ω} = 12 \sec \end{aligned}$

$x = A = A \sin (ω t)$

$ω t = (\frac{π}{2})$

$t = (\frac{π}{2 ω}) = 6$ second

time $= 6 - 2 = 4$ seconds

5. (JEE Main 2023 (Online) 30th January Evening Shift )

The velocity of a particle executing SHM varies with displacement $(x)$ as $4 v^{2} = 50 - x^{2}$ . The time period of oscillations is $\frac{x}{7} s$ . The value of $x$ is ___________. $($ Take $π = \frac{22}{7})$

Correct Answer is (88)

$4 v^{2} = 50 - x^{2}$

or $v = \frac{1}{2} \sqrt{50 - x^{2}}$

Comparing the above equation with $v = ω \sqrt{A^{2} - x^{2}}$

$\Rightarrow ω = \frac{1}{2}$

& $A = \sqrt{50}$

so $\frac{2 π}{T} = \frac{1}{2}$

$\Rightarrow T = 4 π \sec$

$= 4 \times \frac{22}{7} \sec$

$T = \frac{88}{7} \sec$

so $x = 88$