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

Q 1 :

The displacement of a particle executing SHM is given by $x = 10$ $\sin (ω t + \frac{π}{3}) m$ . The time period of motion is 3.14s. The velocity of the particle at $t = 0$ is _______ m/s. [2024]

(10) $x = 10 \sin (ω t + \frac{π}{3})$

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

$at t = 0, x = 10 s i n \frac{π}{3} = 5 \sqrt{3} m$

$n o w v = ω \sqrt{(A^{2} - x^{2})} = 2 \sqrt{10^{2} - {(5 \sqrt{3})}^{2}}$

$v = 2 \sqrt{100 - 75} = 10 m / s$

Q 2 :

A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is ________ cm/s. [2024]

(12) $Given : A = 0.06 cm$

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

$V_{\max} = ω A = 2 \times \frac{6}{100} = 12 cm / s$

Q 3 :

An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of $(\frac{25}{π}) H z$ . At the position $x = 0.04 m$ the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is ________ cm. [2024]

(6) $Total energy = K . E . + P . E .$

$x = 0.04 m,$ $T . E . = 0.5 + 0.4 = 0.9 J$

$\Rightarrow \frac{1}{2} \times 0.2 {(2 π \times \frac{25}{π})}^{2} \times A^{2} = 0.9$

$\Rightarrow A = 0.06 m = 6 c m$

Q 4 :

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 m,$ $2$ $m s^{- 1}$ and $16$ $m s^{- 2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} m$ where $x$ is ________. [2024]

(17) $x = 4 m, v = 2 m / s, a = 16 {m / s}^{2}$

$| a | = ω^{2} x$

$\Rightarrow 16 = ω^{2} (4)$

$ω = 2 rad / s$

$v = ω \sqrt{A^{2} - x^{2}}$

$A = \sqrt{\frac{v^{2}}{ω^{2}} + x^{2}} \Rightarrow A = \sqrt{\frac{4}{4} + 16} = \sqrt{17} m$

Q 5 :

A particle of mass 0.50 kg executes simple harmonic motion under force $F = - 50 (N m^{- 1}) x$ . The time period of oscillation is $\frac{x}{35} s$ . The value of $x$ is _______ $(G i v e n π = \frac{22}{7})$ [2024]

(22) Time period $T = 2 π \sqrt{\frac{m}{k}} = 2 \times \frac{22}{7} \times \sqrt{\frac{0.50}{50}} = \frac{22}{35} s$

On comparing $\frac{x}{35} = \frac{22}{35} \Rightarrow x = 22$

Q 6 :

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is $\sqrt{α}$ cm, where $α =$ ______ . [2024]

(12) $v_{m e a n} = A ω \Rightarrow 10 = 4 ω$

which gives, $ω = \frac{5}{2} r a d / s$

$v = ω \sqrt{A^{2} - x^{2}}$

$5 = \frac{5}{2} \sqrt{4^{2} - x^{2}} \Rightarrow x^{2} = 16 - 4 = 12$

$\Rightarrow x = \sqrt{12} c m$

Q 7 :

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x/8, where x = __________. [2024]

(9) $E = \frac{1}{2} K A^{2}$

$U = \frac{1}{2} K {(\frac{A}{3})}^{2} = \frac{K A^{2}}{2 \times 9} = \frac{E}{9}$

$K E = E - \frac{E}{9} = \frac{8 E}{9}$

Ratio = $\frac{E}{K E} = \frac{E}{\frac{8 E}{9}} = \frac{9}{8} \Rightarrow x = 9$

Q 8 :

A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2A/3. The new amplitude of motion is $n A / 3$ . The value of $n$ is ______ . [2024]

(7) $v = ω \sqrt{A^{2} - x^{2}}$

at $x = \frac{2 A}{3}$

$v = ω \sqrt{A^{2} - {(\frac{2 A}{3})}^{2}} = \frac{\sqrt{5} A ω}{3}$

New amplitude = $A^{'}$

$V^{'} = 3 v = \sqrt{5} A ω = ω \sqrt{{(A^{'})}^{2} - {(\frac{2 A}{3})}^{2}}$

$A^{'} = \frac{7 A}{3}$

Q 9 :

A simple harmonic oscillator has an amplitude A and time period $6 π$ seconds. 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 $\frac{π}{x} s,$ where $x =$ _________ . [2024]

(2)

From phasor diagram particle has to move from P to Q in a circle of radius equal to the amplitude of SHM.

$\cos ϕ = \frac{\frac{\sqrt{3} A}{2}}{A} = \frac{\sqrt{3}}{2} \Rightarrow ϕ = \frac{π}{6}$

$Now, \frac{π}{6} = ω t = \frac{2 π}{T} t$

Q 10 :

A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then $\frac{D}{d}$ is [2025]

$\frac{15}{4}$
25
10
$\frac{16}{5}$

1

2

3

4

(2)

Time period,

In 12.5 s, 6 total oscillations and 1 quarter oscillation.

So, D = (6 $\times$ 4A) + A = 25A

d = A

$\Rightarrow \frac{D}{d} = 25$

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]

1

2

3

4

(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).

1

2

3

4

(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

1

2

3

4

(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

1

2

3

4

(3)

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

1

2

3

4

(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]

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

1

2

3

4

(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]

1

2

3

4

(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

1

2

3

4

(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]

1

2

3

4

(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]

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

1

2

3

4

(3)

$K.E. = P.E.$

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

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

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

Q 21 :

In a linear simple harmonic motion (SHM) [2023]

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

1

2

3

4

(1)

$F = - k x$ A true

$a = - ω^{2} x$ B true

$Velocity is maximum at mean position$ C true

$Acceleration is maximum at extreme points$ D false

Q 22 :

A particle of mass 250 g executes a simple harmonic motion under a periodic force $F = (- 25 x) N$ . The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ________cm. [2023]

(40)

$\frac{1}{4} a = - 25 x \Rightarrow a = - 100 x$

$ω^{2} = 100 \Rightarrow ω = 10$

$ω A = 4 \Rightarrow A = \frac{4}{10} = 0.4 m$

$A = 40 cm$

Q 23 :

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 ________. [2023]

(8)

$x = A \sin (ω t)$

$U_{(x)} = \frac{1}{2} k x^{2}$

$\frac{d U}{d t} = \frac{1}{2} k 2 x \frac{d x}{d t}$

$= k A^{2} ω \sin (ω t) \cos (ω t) \times \frac{2}{2}$

${(\frac{d U}{d t})}_{\max} = \frac{k A^{2} ω}{2} {(\sin 2 ω t)}_{\max}$

$2 ω t = \frac{π}{2} \Rightarrow t = \frac{π}{4} ω = \frac{T}{8} \Rightarrow β = 8$

Q 24 :

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 $π = \frac{22}{7}$ ) [2023]

(88)

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

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

$T = \frac{2 π}{ω} = 4 π = \frac{88}{7}$

$\Rightarrow x = 88$

Q 25 :

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. [2023]

(2)

$K E = P E + \frac{P E}{4}$

$K E = \frac{5}{4} P E$

$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{5}{4} \times \frac{1}{2} m ω^{2} x^{2}$

$[v = ω \sqrt{A^{2} - x^{2}}]$

$A^{2} - x^{2} = \frac{5}{4} x^{2}$

$\frac{9 x^{2}}{4} = A^{2} \Rightarrow x = \frac{2}{3} A$

$∴$ $x = \frac{2}{3} \times 3 cm \Rightarrow x = 2 cm$

Q 26 :

At a given point of time the value of displacement of a simple harmonic oscillator is given as $y = A \cos (30 °)$ . 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 __________ . [2023]

(4)

$General equation for displacement is given by$

$x = A \sin (ω t + ϕ)$

$At given time \Rightarrow ω t + ϕ = 60 °$

$\Rightarrow x = 40 \times \frac{\sqrt{3}}{2} = 20 \sqrt{3} cm$

$\Rightarrow A = 40 cm$

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

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

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

$\Rightarrow k = 10^{4}$

$\Rightarrow x = 4$

Q 27 :

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is ______ Hz.

$[take π = \frac{22}{7}]$ [2026]

14
176
28
88

1

2

3

4

(1)

$ω = 176 rad/sec$

$f_{k} = \frac{ω}{2 π} = \frac{176}{2 \times 22} \times 7$

$= \frac{176}{44} \times 7$

$= 4 \times 7 = 28 Hz$

$So frequency of oscillator$

$f = \frac{k}{2} = 14 Hz$

Q 28 :

The displacement of a particle, executing simple harmonic motion with time period T, is expressed as $x (t) = A \sin ω t$ where A is the amplitude. The maximum value of potential energy of this oscillator is found at $t = \frac{T}{2 β} .$ The value of $β$ is ______. [2026]

(2)

Potential energy is maximum at extreme position. The particle starting at mean position reaches extreme position in time $\frac{T}{4}$ .

Topic Question Set

Q 1 :

The displacement of a particle executing SHM is given by $x = 10$ $\sin (ω t + \frac{π}{3}) m$ . The time period of motion is 3.14s. The velocity of the particle at $t = 0$ is _______ m/s. [2024]

Q 2 :

A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is ________ cm/s. [2024]

Q 3 :

An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of $(\frac{25}{π}) H z$ . At the position $x = 0.04 m$ the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is ________ cm. [2024]

Q 4 :

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 m,$ $2$ $m s^{- 1}$ and $16$ $m s^{- 2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} m$ where $x$ is ________. [2024]

Q 5 :

A particle of mass 0.50 kg executes simple harmonic motion under force $F = - 50 (N m^{- 1}) x$ . The time period of oscillation is $\frac{x}{35} s$ . The value of $x$ is _______ $(G i v e n π = \frac{22}{7})$ [2024]

Q 6 :

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is $\sqrt{α}$ cm, where $α =$ ______ . [2024]

Q 7 :

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x/8, where x = __________. [2024]

Q 8 :

A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2A/3. The new amplitude of motion is $n A / 3$ . The value of $n$ is ______ . [2024]

Q 9 :

A simple harmonic oscillator has an amplitude A and time period $6 π$ seconds. 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 $\frac{π}{x} s,$ where $x =$ _________ . [2024]

Q 10 :

A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then $\frac{D}{d}$ is [2025]

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]

Q 22 :

A particle of mass 250 g executes a simple harmonic motion under a periodic force $F = (- 25 x) N$ . The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ________cm. [2023]

Q 23 :

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 ________. [2023]

Q 24 :

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 $π = \frac{22}{7}$ ) [2023]

Q 25 :

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. [2023]

Q 26 :

At a given point of time the value of displacement of a simple harmonic oscillator is given as $y = A \cos (30 °)$ . 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 __________ . [2023]

Q 27 :

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is ______ Hz.

$[take π = \frac{22}{7}]$ [2026]

Q 28 :

The displacement of a particle, executing simple harmonic motion with time period T, is expressed as $x (t) = A \sin ω t$ where A is the amplitude. The maximum value of potential energy of this oscillator is found at $t = \frac{T}{2 β} .$ The value of $β$ is ______. [2026]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : The displacement of a particle executing SHM is given by x=10 sin(ωt+π3)m. The time period of motion is 3.14s. The velocity of the particle at t=0 is _______ m/s. [2024]

Q 2 : A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is ________ cm/s. [2024]

Q 3 : An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of (25π)Hz. At the position x=0.04m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is ________ cm. [2024]

Q 4 : The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of 4m, 2 ms-1 and 16 ms-2 at a certain instant. The amplitude of the motion is xm where x is ________. [2024]

Q 5 : A particle of mass 0.50 kg executes simple harmonic motion under force F=-50(Nm-1)x. The time period of oscillation is x35s. The value of x is _______ (Given π=227) [2024]

Q 6 : A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is α cm, where α= ______ . [2024]

Q 7 : When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x/8, where x = __________. [2024]

Q 8 : A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2A/3. The new amplitude of motion is nA/3. The value of n is ______ . [2024]

Q 9 : A simple harmonic oscillator has an amplitude A and time period 6π seconds. Assuming the oscillation starts from its mean position, the time required by it to travel from x=A to x=32A will be πxs, where x= _________ . [2024]

Q 10 : A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then Dd is [2025]

Q 11 : A particle oscillates along the x-axis according to the law, x(t)=x0sin2(t2) where x0=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 A2 is 2 s, then the time A taken by the particle in going from x=A2 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 A2, 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=A2 and it moves along the positive x-axis. The displacement of the particle is given by x=Asin(ωt+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]

Q 22 : A particle of mass 250 g executes a simple harmonic motion under a periodic force F=(-25x) N. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ________cm. [2023]

Q 23 : The general displacement of a simple harmonic oscillator is x=Asinωt. Let T be its time period. The slope of its potential energy (U) – time (t) curve will be maximum when t=Tβ. The value of β is ________. [2023]

Q 24 : The velocity of a particle executing SHM varies with displacement (x) as 4v2=50-x2. The time period of oscillations is x7 s. The value of x is _______. (Take π=227) [2023]

Q 25 : 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. [2023]

Q 26 : At a given point of time the value of displacement of a simple harmonic oscillator is given as y=Acos(30°). If amplitude is 40 cm and kinetic energy at that time is 200 J, the value of force constant is 1.0×10x Nm-1. The value of x is __________ . [2023]

Q 27 : The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is ______ Hz. [take π=227] [2026]

Q 28 : The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t)=Asinωt where A is the amplitude. The maximum value of potential energy of this oscillator is found at t=T2β. The value of β is ______. [2026]

Want Us to Email you About Special Offers & Updates?

Q 1 :

The displacement of a particle executing SHM is given by $x = 10$ $\sin (ω t + \frac{π}{3}) m$ . The time period of motion is 3.14s. The velocity of the particle at $t = 0$ is _______ m/s. [2024]

Q 2 :

A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is ________ cm/s. [2024]

Q 3 :

An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of $(\frac{25}{π}) H z$ . At the position $x = 0.04 m$ the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is ________ cm. [2024]

Q 4 :

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 m,$ $2$ $m s^{- 1}$ and $16$ $m s^{- 2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} m$ where $x$ is ________. [2024]

Q 5 :

A particle of mass 0.50 kg executes simple harmonic motion under force $F = - 50 (N m^{- 1}) x$ . The time period of oscillation is $\frac{x}{35} s$ . The value of $x$ is _______ $(G i v e n π = \frac{22}{7})$ [2024]

Q 6 :

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is $\sqrt{α}$ cm, where $α =$ ______ . [2024]

Q 7 :

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x/8, where x = __________. [2024]

Q 8 :

A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2A/3. The new amplitude of motion is $n A / 3$ . The value of $n$ is ______ . [2024]

Q 9 :

A simple harmonic oscillator has an amplitude A and time period $6 π$ seconds. 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 $\frac{π}{x} s,$ where $x =$ _________ . [2024]

Q 10 :

A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then $\frac{D}{d}$ is [2025]

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]

Q 22 :

A particle of mass 250 g executes a simple harmonic motion under a periodic force $F = (- 25 x) N$ . The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ________cm. [2023]

Q 23 :

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 ________. [2023]

Q 24 :

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 $π = \frac{22}{7}$ ) [2023]

Q 25 :

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. [2023]

Q 26 :

At a given point of time the value of displacement of a simple harmonic oscillator is given as $y = A \cos (30 °)$ . 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 __________ . [2023]

Q 27 :

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is ______ Hz.

$[take π = \frac{22}{7}]$ [2026]

Q 28 :

The displacement of a particle, executing simple harmonic motion with time period T, is expressed as $x (t) = A \sin ω t$ where A is the amplitude. The maximum value of potential energy of this oscillator is found at $t = \frac{T}{2 β} .$ The value of $β$ is ______. [2026]