A particle is executing a simple harmonic motion of period 2 s. When it is at its extreme displacement from its mean position, it receives an additional energy equal to what it had in its mean position. Due to this in its subsequent motion:

Easy

A

its amplitude will change and become equal to $\sqrt{2}$ times its previous amplitude
B

its periodic time will become doubled, i.e.,4 s
C

its potential energy will be decreased
D

it will continue to execute simple harmonic motion of the same amplitude and period as before receiving the additional energy

Solution

$E' = \frac{1}{2} {mω}^{2} d^{2} = {mω}^{2} a^{2} \Rightarrow a' = \sqrt{2} a$

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