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Question

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance 2A3 from equilibrium position. The new amplitude of the motion is.
  1. A341
  2. 3A
  3. A3
  4. 7A3

A
3A
B
A3
C
7A3
D
A341
Solution
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In SHM, V=A2x2

V=A24A29=5A29

3V=A24A29

Dividing the above two equations give A=7A3

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