Question

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 the equilibrium position. The new amplitude of motion is
A. $\dfrac{A}{3}\sqrt {41}$
B. $3A$
C. $A\sqrt 3$
D. $\dfrac{{7A}}{3}$

Answer 1

A special type of periodic motion where the restoring force of the moving object is directly proportional to its displacement magnitude and which acts towards the object's equilibrium position is called simple harmonic motion. It is the periodic motion of a point along a straight line such that its acceleration is always towards a fixed point, and it is directly proportional to its distance from that point. In this question, first, find the velocity of the particle with amplitude A and then find the change in amplitude when the velocity is tripled at a distance $\dfrac{{2A}}{3}$ from the equilibrium position.

Complete Step by Step Answer:
Let, the displacement function be $x = A\sin \omega t$
Where,
$A$ is the amplitude
$\omega$ is the angular frequency$= \dfrac{{2\pi }}{T}$
$T$ is time period
The velocity in simple harmonic motion is given as
$v = \omega \sqrt {{A^2} - {x^2}} ………..(i)$
When the distance $x = \dfrac{{2A}}{3}$, the velocity becomes 3v, hence
$3v = \omega \sqrt {{{A'}^2} - {{\left( {\dfrac{{2A}}{3}} \right)}^2}} ……….. (ii)$
For the equation (i), we need to find the velocity of the particle if it is at $x = \dfrac{{2A}}{3}$; hence we can write equation (i) as
$v = \omega \sqrt {{A^2} - {{\left( {\dfrac{{2A}}{3}} \right)}^2}} ……….. (iii)$
Now divide equation (ii) by (iii) and solve.

$$\dfrac{{3v}}{v} = \dfrac{{\omega \sqrt {{{A'}^2} - {{\left( {\dfrac{{2A}}{3}} \right)}^2}} }}{{\omega \sqrt {{A^2} - {{\left( {\dfrac{{2A}}{3}} \right)}^2}} }} \\\ \dfrac{3}{1} = \dfrac{{\sqrt {{{A'}^2} - {{\left( {\dfrac{{2A}}{3}} \right)}^2}} }}{{\sqrt {{A^2} - {{\left( {\dfrac{{2A}}{3}} \right)}^2}} }} \\\ \dfrac{9}{1} = \dfrac{{{{A'}^2} - {{\left( {\dfrac{{2A}}{3}} \right)}^2}}}{{{A^2} - {{\left( {\dfrac{{2A}}{3}} \right)}^2}}} \\\$$

By cross multiplying,

$$\\\ 9\left( {{A^2} - \dfrac{{4{A^2}}}{9}} \right) = {{A'}^2} - \dfrac{{4{A^2}}}{9} \\\ 9\left( {\dfrac{{9{A^2} - 4{A^2}}}{9}} \right) = \dfrac{{9{{A'}^2} - 4{A^2}}}{9} \\\ 9 \times 5{A^2} = 9{{A'}^2} - 4{A^2} \\\ \therefore 9{{A'}^2} = 45{A^2} + 4{A^2} \\\ {{A'}^2} = \dfrac{{49{A^2}}}{9} \\\ A' = \dfrac{{7A}}{3} \\\$$

Hence, option D is correct.

Note: Simple Harmonic Motion, abbreviated as SHM is basically repetitive movement back and forth of an object through an equilibrium or central position such that maximum displacement on one side of its position is equal to maximum displacement on the other side.