
A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance $\dfrac{{2A}}{3}$ from equilibrium position. The new amplitude of the motion is:
A) $\dfrac{A}{3}\sqrt {41} $
B) $3A$
C) $A\sqrt 3 $
D) $\dfrac{{7A}}{3}$
Answer
232.8k+ views
Hint: Recall that the simple harmonic motion is defined as the motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of the restoring force is towards its mean position.
Complete step by step solution:
The velocity of a particle executing SHM at any instant, is the time rate of change of its displacement at that instant.
$v = \omega \sqrt {{A^2} - {x^2}} $
Where $\omega $is the angular frequency, A is the amplitude and x is the displacement of the particle.
Suppose that the new amplitude of the motion be ‘A’.
If A is the initial amplitude and$\omega $is the angular frequency. Initial velocity of a particle performs SHM,
$\Rightarrow {v^2} = {\omega ^2}[{A^2} - {(\dfrac{{2A}}{3})^2}]$---(i)
The final velocity of the particle is given by
$\Rightarrow {(3v)^2} = {\omega ^2}[A{'^2} - {(\dfrac{{2A}}{3})^2}]$---(ii)
The ratio of equation (i) and (ii) is written as,
$\Rightarrow \dfrac{{{v^2}}}{{{{(3v)}^2}}} = \dfrac{{{A^2} - \dfrac{{4{A^2}}}{9}}}{{A{'^2} - \dfrac{{4{A^2}}}{9}}}$
\[ \Rightarrow \dfrac{1}{9} = \dfrac{{\dfrac{{9{A^2} - 4{A^2}}}{9}}}{{A{'^2} - \dfrac{{4{A^2}}}{9}}}\]
$ \Rightarrow \dfrac{1}{9} = \dfrac{{\dfrac{{5{A^2}}}{9}}}{{A{'^2} - \dfrac{{4{A^2}}}{9}}}$
\[ \Rightarrow \dfrac{1}{9}(A{'^2} - \dfrac{{4{A^2}}}{9}) = \dfrac{{5{A^2}}}{9}\]
$ \Rightarrow {A' ^2} - \dfrac{{4{A^2}}}{9} = \dfrac{{5{A^2}}}{9} \times 9$
$ \Rightarrow {A' ^2} = 5{A^2} + \dfrac{{4{A^2}}}{9}$
$\Rightarrow {A' ^2} = \dfrac{{45{A^2} + 4{A^2}}}{9}$
$\Rightarrow {A’ ^2} = \dfrac{{49{A^2}}}{9}$
$\Rightarrow A' = \dfrac{{7A}}{3}$
The new amplitude of the motion is $\dfrac{{7A}}{3}$.
Option D is the right answer.
Note: It is important to note that all the simple harmonic motions are periodic but all periodic motions are not simple harmonic. This is because a simple harmonic motion is a type of periodic motion in which there is to and fro movement of an object about its mean position. But the mean position of the object can be different. But periodic motion is a type of motion which repeats itself after fixed intervals of time. For example the motion of Earth around the sun is a type of periodic motion as it repeats after a fixed interval of time.
Complete step by step solution:
The velocity of a particle executing SHM at any instant, is the time rate of change of its displacement at that instant.
$v = \omega \sqrt {{A^2} - {x^2}} $
Where $\omega $is the angular frequency, A is the amplitude and x is the displacement of the particle.
Suppose that the new amplitude of the motion be ‘A’.
If A is the initial amplitude and$\omega $is the angular frequency. Initial velocity of a particle performs SHM,
$\Rightarrow {v^2} = {\omega ^2}[{A^2} - {(\dfrac{{2A}}{3})^2}]$---(i)
The final velocity of the particle is given by
$\Rightarrow {(3v)^2} = {\omega ^2}[A{'^2} - {(\dfrac{{2A}}{3})^2}]$---(ii)
The ratio of equation (i) and (ii) is written as,
$\Rightarrow \dfrac{{{v^2}}}{{{{(3v)}^2}}} = \dfrac{{{A^2} - \dfrac{{4{A^2}}}{9}}}{{A{'^2} - \dfrac{{4{A^2}}}{9}}}$
\[ \Rightarrow \dfrac{1}{9} = \dfrac{{\dfrac{{9{A^2} - 4{A^2}}}{9}}}{{A{'^2} - \dfrac{{4{A^2}}}{9}}}\]
$ \Rightarrow \dfrac{1}{9} = \dfrac{{\dfrac{{5{A^2}}}{9}}}{{A{'^2} - \dfrac{{4{A^2}}}{9}}}$
\[ \Rightarrow \dfrac{1}{9}(A{'^2} - \dfrac{{4{A^2}}}{9}) = \dfrac{{5{A^2}}}{9}\]
$ \Rightarrow {A' ^2} - \dfrac{{4{A^2}}}{9} = \dfrac{{5{A^2}}}{9} \times 9$
$ \Rightarrow {A' ^2} = 5{A^2} + \dfrac{{4{A^2}}}{9}$
$\Rightarrow {A' ^2} = \dfrac{{45{A^2} + 4{A^2}}}{9}$
$\Rightarrow {A’ ^2} = \dfrac{{49{A^2}}}{9}$
$\Rightarrow A' = \dfrac{{7A}}{3}$
The new amplitude of the motion is $\dfrac{{7A}}{3}$.
Option D is the right answer.
Note: It is important to note that all the simple harmonic motions are periodic but all periodic motions are not simple harmonic. This is because a simple harmonic motion is a type of periodic motion in which there is to and fro movement of an object about its mean position. But the mean position of the object can be different. But periodic motion is a type of motion which repeats itself after fixed intervals of time. For example the motion of Earth around the sun is a type of periodic motion as it repeats after a fixed interval of time.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

