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A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2A3. The new amplitude of motion is nA3. The value of n is .
7
v= A2-x2 ( is unchanged; it depends only on the system)
At x= 2A3: v= A2- 4A29= 53 A
New speed: 3v= 5 A= A'2- 4A29
A'2=5A2+ 4A29= 49A29
A'= 7A3 ⇒ n = 7
Answer: 7
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