JEE Main SHM PYQ

6. (JEE Main 2023 (Online) 13th April Morning Shift )

At a given point of time the value of displacement of a simple harmonic oscillator is given as $y = A \cos (30^{\circ})$ . If amplitude is $40 cm$ and kinetic energy at that time is $200 J$ , the value of force constant is $1.0 \times 10^{x} {Nm}^{- 1}$ . The value of $x$ is ____________.

Correct Answer is (4)

Given the general equation for displacement in a simple harmonic oscillator:

$x = A \sin (ω t + ϕ)$

At the given time, we have:

$ω t + ϕ = 30^{\circ}$

Given the amplitude $A = 40 cm$ and the displacement $x = 40 \times \frac{\sqrt{3}}{2} cm = 20 \sqrt{3} cm$ , we can write the kinetic energy, $K E$ , as:

$K E = \frac{1}{2} k (A^{2} - x^{2}) = 200$

Now, substitute the values for $A$ and $x$ :

$200 = \frac{1}{2} k (\frac{1600 - 1200}{100 \times 100})$

Simplify the equation:

$400 \times 100 \times 100 = k \times 400$

Solve for the force constant, $k$ :

$k = 10^{4} {Nm}^{- 1}$

Given that the force constant is expressed as $k = 1.0 \times 10^{x} {Nm}^{- 1}$ , comparing the values, we get:

$1.0 \times 10^{x} = 10^{4}$

Thus, the value of $x$ is $4$ .

7. (JEE Main 2023 (Online) 1st February Morning Shift )

The amplitude of a particle executing SHM is $3 cm$ . The displacement at which its kinetic energy will be $25 %$ more than the potential energy is: __________ $cm$

Correct Answer is (2)

Given the amplitude $A$ and the length $l$ of the pendulum, we can find the maximum angular displacement $θ_{0}$ :

$\sin θ_{0} = \frac{A}{l} = \frac{10}{100} = \frac{1}{10}$

By conservation of energy, the following equation holds:

$\frac{1}{2} m v^{2} = m g l (1 - \cos θ)$

The maximum tension occurs at the mean position (i.e., when the pendulum is vertical). At this point, we have:

$\begin{aligned} T - m g = \frac{m v^{2}}{l} & \Rightarrow T = m g + \frac{m v^{2}}{l} \end{aligned}$

Substituting the conservation of energy equation, we get:

$\begin{aligned} T = m g + 2 m g (1 - \cos θ) \\ = m g [1 + 2 (1 - \sqrt{1 - \sin^{2} θ})] \\ = m g [3 - 2 \sqrt{1 - \frac{1}{100}}] \\ = \frac{250}{1000} \times 10 [3 - 2 (1 - \frac{1}{200})] = \frac{99}{40} \\ ∴ x = 99 \end{aligned}$

So, the value of $x$ is 99.

8. (JEE Main 2023 (Online) 30th January Morning Shift )

The general displacement of a simple harmonic oscillator is $x = A \sin ω t$ . Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $t = \frac{T}{β}$ . The value of $β$ is ______________.

Correct Answer is (8)

$U = \frac{1}{2} m ω^{2} A^{2} \sin^{2} ω t$

So, $\frac{d U}{d t} = \frac{m ω^{3} A^{2}}{2} \sin 2 ω t$

This value will be maximum when $\sin 2 ω t = 1$

or $2 ω t = \frac{π}{2}$

$2 \times \frac{2 π}{T} t = \frac{π}{2}$

$\Rightarrow t = \frac{T}{8}$

So $β = 8$

9. (JEE Main 2022 (Online) 28th July Evening Shift )

The potential energy of a particle of mass $4 kg$ in motion along the x-axis is given by $U = 4 (1 - \cos 4 x)$ J. The time period of the particle for small oscillation $(\sin θ ≃ θ)$ is $(\frac{π}{K}) s$ . The value of $K$ is _________.

Correct Answer is (2)

$U = 4 (1 - \cos 4 x)$

$\Rightarrow F = - \frac{d U}{d x} = - (4) (4 \sin 4 x)$

$= - 16 \sin 4 x$

as small x

$F = - 16 (4 x) = - 64 x \equiv - k x$

$T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{4}{64}} = \frac{π}{2}$

$\Rightarrow K = 2$

10. (JEE Main 2021 (Online) 31st August Evening Shift)

For a body executing S.H.M. :

(1) Potential energy is always equal to its K.E.

(2) Average potential and kinetic energy over any given time interval are always equal.

(3) Sum of the kinetic and potential energy at any point of time is constant.

(4) Average K.E. in one time period is equal to average potential energy in one time period.

Choose the most appropriate option from the options given below :

A. (3) and (4)

B. only (3)

C. (2) and (3)

D. only (2)

Correct Answer is Option (A)

In S.H.M. total mechanical energy remains constant and also = = $\frac{1}{4}$ KA² (for 1 time period)

Simple Harmonic Motion