First slide

Simple harmonic motion

Question

A particle performs simple harmonic motion with amplitude A. It’s speed is trebled at the instant that it is a distance 2A/3 from equilibrium position. The new amplitude of the motion is

Moderate

A

$\frac{A}{3} \sqrt{41}$
B

$3 A$
C

$A \sqrt{3}$
D

$\frac{7}{3} A$

Solution

initial velocity of a particle performs SHM
$v=w \sqrt{A^{2} {-x}^{2}}$ given $x= \frac{2}{3} A$
$v=w \sqrt{A^{2} - {(2A/3)}^{2}}$
$v^{2} {=w}^{2} [A^{2} - {(\frac{2A}{3})}^{2}]$ ------(1)
Where A is initial amplitude, w is angular frequency
Final velocity
${(3 v)}^{2} = w^{2} [{(A^{1})}^{2} - {(\frac{2 A}{3})}^{2}]$ -----(2)
From equation(1) and (2) we get
$\frac{1}{9} = \frac{A^{2} - \frac{{4A}^{2}}{9}}{{{(A}^{1})}^{2} - \frac{{4A}^{2}}{9}} {{(A}^{1})}^{2} - \frac{{4A}^{2}}{9} =9 [A^{2} - \frac{{4A}^{2}}{9}] {=9A}^{2} -9 \frac{{4A}^{2}}{9} {{(A}^{1})}^{2} =5 A^{2} + \frac{{4A}^{2}}{9} = \frac{{49A}^{2}}{9} A^{1} = \frac{{7A}^{}}{3}$ .
new Amplitude is $A^{1}$

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