JEE Main SHM PYQ

1. (JEE Main 2023 (Online) 6th April Morning Shift)

A mass $m$ is attached to two strings as shown in figure. The spring constants of two springs are $K_{1}$ and $K_{2}$ . For the frictionless surface, the time period of oscillation of mass $m$ is :

JEE Main 2023 (Online) 6th April Morning Shift Physics - Simple Harmonic Motion Question 2 English

A. $2 π \sqrt{\frac{m}{K_{1} + K_{2}}}$

B. $2 π \sqrt{\frac{m}{K_{1} - K_{2}}}$

C. $\frac{1}{2 π} \sqrt{\frac{K_{1} + K_{2}}{m}}$

D. $\frac{1}{2 π} \sqrt{\frac{K_{1} - K_{2}}{m}}$

Correct Answer is Option (A)

Since, both spring are in parallel

Combination, therefore, $K_{eq} = K_{1} + K_{2}$

Time period of oscillation, $T = 2 π \sqrt{\frac{m}{K_{eq}}}$

$\Rightarrow T = 2 π \sqrt{\frac{m}{K_{1} + K_{2}}}$

Option (a) is correct.

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

For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 kg$ , the angular frequency is $ω_{1}$ . When the mass block is $2 kg$ the angular frequency is $ω_{2}$ . The ratio $ω_{2} / ω_{1}$ is

JEE Main 2023 (Online) 30th January Evening Shift Physics - Simple Harmonic Motion Question 19 English

A. $1 / \sqrt{2}$

B. $1 / 2$

C. 2

D. $\sqrt{2}$

Correct Answer is Option (A)

$ω = \sqrt{\frac{K}{m}} \Rightarrow ω \propto \frac{1}{\sqrt{m}}$

$\frac{ω_{2}}{ω_{1}} = \sqrt{\frac{m_{1}}{m_{2}}} = \sqrt{\frac{1}{2}}$

3.(JEE Main 2021 (Online) 18th March Morning Shift )

A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a}$ s. The value of 'a' to the nearest integer is _________.

Correct Answer is (6)

Time period (T) = 2 sec.

X = A sin ( $ω$ t + $ϕ$ ) ( $ϕ$ = 0 at M.P.)

$\Rightarrow$ $\frac{A}{2} = A \sin \frac{2 π}{T} t$

$\Rightarrow$ $\frac{2 π}{2} t = \frac{π}{6}$

$\Rightarrow$ $t = \frac{1}{6}$

$∴$ a = 6

4.(JEE Main 2023 (Online) 10th April Evening Shift )

A rectangular block of mass $5 kg$ attached to a horizontal spiral spring executes simple harmonic motion of amplitude $1 m$ and time period $3.14 s$ . The maximum force exerted by spring on block is _________ N

Correct Answer is (20)

To find the maximum force exerted by the spring on the block, we can use Hooke's law and the properties of simple harmonic motion.

First, let's find the angular frequency $ω$ :

$ω = \frac{2 π}{T}$

where $T = 3.14 s$ is the time period.

$ω = \frac{2 π}{3.14} \approx 2 rad / s$

Now, let's find the maximum velocity $v_{m a x}$ of the block:

$v_{m a x} = ω A$

where $A = 1 m$ is the amplitude.

$v_{m a x} = 2 rad / s \times 1 m = 2 m / s$

Next, we can find the spring constant $k$ using the mass of the block $m = 5 kg$ and the angular frequency $ω$ :

$ω^{2} = \frac{k}{m} \Rightarrow k = m ω^{2}$

$k = 5 kg \times (2 rad / s)^{2} = 20 N / m$

Finally, we can find the maximum force exerted by the spring on the block. At maximum displacement, the force is given by Hooke's law:

$F_{m a x} = k A$

$F_{m a x} = 20 N / m \times 1 m = 20 N$

The maximum force exerted by the spring on the block is $20 N$ .

5.(JEE Main 2023 (Online) 31st January Morning Shift )

In the figure given below, a block of mass $M = 490 g$ placed on a frictionless table is connected with two springs having same spring constant $(K = 2 N m^{- 1})$ . If the block is horizontally displaced through $X$ ' $m$ then the number of complete oscillations it will make in $14 π$ seconds will be _____________.

JEE Main 2023 (Online) 31st January Morning Shift Physics - Simple Harmonic Motion Question 20 English

Correct Answer is (20)

$K_{eff} = K + K$ $= 2 k$ (As both springs are in use in parallel)

$= 2 \times 2 = 4 N / m$

$m = 490 gm$

$\begin{aligned} = 0.49 kg \end{aligned}$

$T = 2 π \sqrt{\frac{m}{Keff}} = 2 π \sqrt{\frac{0.49 kg}{4}}$

$= 2 π \sqrt{\frac{49}{400}} = 2 π \frac{7}{20} = \frac{7 π}{10}$

No. of oscillation in the $14 π$ is

$N = \frac{time}{T} = \frac{14 π}{7 π / 10} = 20$