Solution:
If equation of motion in shm is x = A sin ωt then eqn. of potential energy = ½k x²= ½kA² sin²ωt
\[ U = \frac{1}{2}K A^{2}\left( \frac{1- sin(2\omega t)}{2} \right) \]
Frequency becomes double so time period becomes half.
A particle executes an undamped SHM of time period T. Then the time period with which the potential energy changes is :
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