A particle performs SHM with amplitude A. its speed is trebled at the instant when it is at a distance of 2A3from equilibrium position. The new amplitude of the motion is

V=ωA2−x2→(1) A=Inital amplitude 3V=ωA′2−x2→(2)A′=new amplitude (2)(1),3=A2−x2A2−x2 then A′2−x2=9A2−9x2 A′2=9A2−8x2=9A2−82A32=9A2−329A2=499A2 Then A1=7A3

A particle performs SHM with amplitude A. its speed is trebled at the instant when it is at a distance of 2A3from equilibrium position. The new amplitude of the motion is

V=ωA2−x2→(1) A=Inital amplitude 3V=ωA′2−x2→(2)A′=new amplitude (2)(1),3=A2−x2A2−x2 then A′2−x2=9A2−9x2 A′2=9A2−8x2=9A2−82A32=9A2−329A2=499A2 Then A1=7A3

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A particle performs SHM with amplitude A. its speed is trebled at the instant when it is at a distance of $\frac{2 A}{3}$ from equilibrium position. The new amplitude of the motion is

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$\frac{5 A}{3}$

$\frac{4 A}{3}$

$\frac{7 A}{3}$

$\frac{8 A}{3}$

answer is C.

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Detailed Solution

$V = ω \sqrt{A^{2} - x^{2}} \to (1) A = Inital amplitude 3 V = ω \sqrt{A^{' 2} - x^{2}} \to (2) A^{'} = new amplitude \frac{(2)}{(1)}, 3 = \frac{\sqrt{A^{2} - x^{2}}}{(A^{2} - x^{2}} then A^{' 2} - x^{2} = 9 A^{2} - 9 x^{2} \begin{array}{l} A^{' 2} = 9 A^{2} - 8 x^{2} = 9 A^{2} - 8 {(\frac{2 A}{3})}^{2} \\ = 9 A^{2} - \frac{32}{9} A^{2} = \frac{49}{9} A^{2} \end{array} Then A^{1} = \frac{7 A}{3}$