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

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]

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(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]

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(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]

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(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]

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(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]

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(4)

$General equation for displacement is given by$

$x = A \sin (ω t + ϕ)$

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

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

$\Rightarrow A = 40 cm$

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

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

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

$\Rightarrow k = 10^{4}$

$\Rightarrow x = 4$

Topic Question Set

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]

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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]

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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]

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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]

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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]

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