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A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A/3  from equilibrium position. The new amplitude of the motion is

(a) A41/3

(b) 3A

(c) A3

(d) 7A/3

1 Answer

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Best answer

Correct Option (d) 7A/3

Explanation:

The velocity of a particle executing SHM at any instant, is defined as the time rate of change of its displacement at that instant.

where, ω is angular frequency, A is amplitude and x is displacement of a particle.

Suppose that the new amplitude of the motion be A'.

Initial velocity of a particle performs SHM,

where, A is initial amplitude and ω is angular frequency,

Final velocity

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