First slide

Applications of SHM

Question

A particle performs simple harmonic motion with amplitude A. It’s speed is trebled at the instant that it is at 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

$initial velocity of a particle performs SHM =v=w \sqrt{A^{2} {-x}^{2}} given x= \frac{2 A}{3};A is initial amplitude, w is angular frequency v = w \sqrt{A^{2} - {(\frac{2 A}{3})}^{2}} s q u a r i n g b o t h s i d e s v^{2} = w^{2} (A^{2} - {(\frac{2 A}{3})}^{2}) - - - (1) s p e e d i s t r e b l e d a t t h e i n s \tan t t h a t i t i s a t a d i s \tan c e 2 A / 3 l e t n e w a m p l i t u d e b e A^{1} {(3 v)}^{2} = w^{2} ({A^{1}}^{2} - {(\frac{2 A}{3})}^{2}) - - - (2) d i v i d i n g e q n (2) w i t h e q n (1) \frac{{(3 v)}^{2}}{v^{2}} = \frac{w^{2} ({A^{1}}^{2} - {(\frac{2 A}{3})}^{2})}{w^{2} (A^{2} - {(\frac{2 A}{3})}^{2})} 9 = \frac{{A^{1}}^{2} - {(\frac{2 A}{3})}^{2}}{A^{2} - {(\frac{2 A}{3})}^{2}} A^{1} = \frac{7 A}{3}$

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