A) \[\frac{7A}{3}\]
B) \[\frac{A}{3}\sqrt{41}\]
C) 3A
D) A 3
Correct Answer: A
Solution :
Let new amplitude is A' initial velocity \[{{v}^{2}}={{\omega }^{2}}\left( {{A}^{2}}-{{\left( \frac{2A}{3} \right)}^{2}} \right)\] ?(1) Where A is initial amplitude & w is angular frequency. Final velocity \[{{(3v)}^{2}}={{\omega }^{2}}\left( A{{'}^{2}}-{{\left( \frac{2A}{3} \right)}^{2}} \right)\] ...(2) From equation & equation (2) \[\frac{1}{9}=\frac{{{A}^{2}}-\frac{4{{A}^{2}}}{9}}{A{{'}^{2}}-\frac{4{{A}^{2}}}{9}}\] \[\]
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