Oscillation and Simple Harmonic Motion | Physics JEE previous year questions with Solutions

Q 11 :

A particle oscillates along the x-axis according to the law, $x (t) = x_{0} \sin^{2} (\frac{t}{2})$ where $x_{0} = 1 m$ . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph. [2025]

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(1)

$x (t) = x_{0} \sin^{2} (\frac{t}{2}) = \frac{x_{0}}{2} (1 - \cos t)$

Clearly $\frac{x_{0}}{2}$ is mean position, Particle is oscillating between

Q 12 :

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 $ω$ .

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: [2025]

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).

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(4)

$x = A \sin (ω t + ϕ)$

$x_{0} = A \sin ϕ$ ... (i)

$p = m A ω \cos (ω t + ϕ)$

$p_{0} = m A ω \cos ϕ$ ... (ii)

(ii)/(i)

$\Rightarrow \tan ϕ = (\frac{x_{0}}{p_{0}}) m ω$

$\sin ϕ = \frac{x_{0} m ω}{\sqrt{{(m ω x_{0})}^{2} + p_{0}^{2}}}$

From (i),

$A = \frac{x_{0}}{\sin ϕ} = \frac{\sqrt{{(m ω x_{0})}^{2} + p_{0}^{2}}}{m ω}$

Hence both position and linear momentum of a particle can be expressed as a function of time if we know initial momentum and position.

Q 13 :

Which of the following curves possibly represent one-dimensional motion of a particle? [2025]

Choose the correct answer from the options given below:

A, B and D only
A, B and C only
A and B only
A, C and D only

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(1)

A. Phase increase with time in SHM, $ϕ$ = kt + C

For example, in SHM, x = A sin $ϕ$

$\Rightarrow$ Correct

B. In SHM Velocity and displacement are related in elliptical/circular relation

i.e., $v^{2} + x^{2}$ = constant, it can be 1 D motion

$\Rightarrow$ Correct

C. At same time particle can't have two velocities $\Rightarrow$ Incorrect.

D. Distance always increases $\Rightarrow$ Correct

Hence A, B and D are correct.

Q 14 :

A particle executes simple harmonic motion between $x = - A$ and $x = + A$ . If the time taken by $A$ the particle to go from $x = 0$ to $\frac{A}{2}$ is $2 s$ , then the time $A$ taken by the particle in going from $x = \frac{A}{2}$ to $A$ is [2023]

1.5 s
3 s
4 s
2 s

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(3)

[IMAGE 99]

Let time from 0 to $\frac{A}{2}$ is $t_{1}$ and from $\frac{A}{2}$ to $A$ is $t_{2}$

Then $ω t_{1} = \frac{π}{6}, ω t_{2} = \frac{π}{3}$

$\frac{t_{1}}{t_{2}} = \frac{1}{2}, t_{2} = 2 t_{1} = 2 \times 2 = 4 sec$

Q 15 :

The maximum potential energy of a block executing simple harmonic motion is 25 J. $A$ is the amplitude of oscillation. At $\frac{A}{2}$ , the kinetic energy of the block is [2023]

37.5 J
9.75 J
18.75 J
12.5 J

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(3)

$u_{\max} = \frac{1}{2} m ω^{2} A^{2} = 25 J$

$KE at \frac{A}{2} = \frac{1}{2} m v_{1}^{2} = \frac{1}{2} m ω^{2} (A^{2} - \frac{A^{2}}{4})$

$KE = \frac{1}{2} m ω^{2} \frac{3 A^{2}}{4} = \frac{3}{4} (\frac{1}{2} m ω^{2} A^{2})$

$KE = \frac{3}{4} \times 25 = 18.75 J$

Q 16 :

A particle executes S.H.M. of amplitude $A$ along the $x$ -axis. At $t = 0$ , the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$ -axis. The displacement of the particle is given by $x = A \sin (ω t + δ),$ then the value of $δ$ will be [2023]

$\frac{π}{6}$
$\frac{π}{3}$
$\frac{π}{4}$
$\frac{π}{2}$

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(1)

$x = A \sin (ω t + δ)$

$v = A ω \cos (ω t + δ)$

$\frac{A}{2} = A \sin (ω t + δ)$

$∴$ $v is +ve$

$At t = 0, in 1st quadrant or 4th quadrant$

$\sin δ = \frac{1}{2} \Rightarrow δ = \frac{π}{6}, \frac{5 π}{6}$ $quadrant$

$∴$ $Common solution is δ = \frac{π}{6}$

Q 17 :

The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement $(x)$ , starting from mean position to extreme position $(A)$ , is given by [2023]

[IMAGE 100]
[IMAGE 101]
[IMAGE 102]
[IMAGE 103]

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(4)

$For a particle executing SHM$

$KE = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$

$When x = 0, KE is maximum and when x = A, KE is zero, and the KE vs x graph is parabolic.$

Q 18 :

A particle is executing Simple Harmonic Motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be [2023]

1 : 1
2 : 1
1 : 4
1 : 3

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(4)

$x = \frac{A}{2}, P.E. = \frac{1}{2} k x^{2}$

$K.E. = \frac{1}{2} k A^{2} - \frac{1}{2} k x^{2}$

$\frac{P.E.}{K.E.} = \frac{x^{2}}{A^{2} - x^{2}} = \frac{A^{2}}{4 (\frac{3 A^{2}}{4})} = \frac{1}{3}$

Q 19 :

Which graph represents the difference between total energy and potential energy of a particle executing SHM vs its distance from mean position? [2023]

[IMAGE 104]
[IMAGE 105]
[IMAGE 106]
[IMAGE 107]

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(4)

$T.E. - P.E. = K.E.$

$K.E. = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$

$Which is the equation of a downward parabola.$

Q 20 :

A particle executes SHM of amplitude $A$ . The distance from the mean position when its kinetic energy becomes equal to its potential energy is _______ [2023]

$\sqrt{2} A$
$2 A$
$\frac{1}{\sqrt{2}} A$
$\frac{1}{2} A$

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(3)

$K.E. = P.E.$

$\frac{1}{2} M ω^{2} (A^{2} - x^{2}) = \frac{1}{2} M ω^{2} x^{2}$

$A^{2} - x^{2} = x^{2} \Rightarrow A^{2} = 2 \times 2$

$\Rightarrow x = \pm \frac{A}{\sqrt{2}}$

Topic Question Set

Q 11 :

A particle oscillates along the x-axis according to the law, $x (t) = x_{0} \sin^{2} (\frac{t}{2})$ where $x_{0} = 1 m$ . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph. [2025]

Q 13 :

Which of the following curves possibly represent one-dimensional motion of a particle? [2025]

Choose the correct answer from the options given below:

Q 14 :

A particle executes simple harmonic motion between $x = - A$ and $x = + A$ . If the time taken by $A$ the particle to go from $x = 0$ to $\frac{A}{2}$ is $2 s$ , then the time $A$ taken by the particle in going from $x = \frac{A}{2}$ to $A$ is [2023]

Q 15 :

The maximum potential energy of a block executing simple harmonic motion is 25 J. $A$ is the amplitude of oscillation. At $\frac{A}{2}$ , the kinetic energy of the block is [2023]

Q 16 :

A particle executes S.H.M. of amplitude $A$ along the $x$ -axis. At $t = 0$ , the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$ -axis. The displacement of the particle is given by $x = A \sin (ω t + δ),$ then the value of $δ$ will be [2023]

Q 17 :

The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement $(x)$ , starting from mean position to extreme position $(A)$ , is given by [2023]

Q 18 :

A particle is executing Simple Harmonic Motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be [2023]

Q 19 :

Which graph represents the difference between total energy and potential energy of a particle executing SHM vs its distance from mean position? [2023]

Q 20 :

A particle executes SHM of amplitude $A$ . The distance from the mean position when its kinetic energy becomes equal to its potential energy is _______ [2023]

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