First slide

Simple hormonic motion

Question

Starting from the origin a body oscillates simple harmonically with period of 2 s. After what time will its kinetic energy be 75% of the total energy

NA

A

1 /6 s
B

1 /4 s
C

1 / 3 s
D

1/12 s

Solution

The kinetic energy of a body undergoing S.H.M. is given by
$K.E. = \frac{1}{2} {mÏ}^{2} A^{2} \cos^{2} â¡ Ït or {(K.E.)}_{max} = \frac{1}{2} {mÏ}^{2} A^{2}$
According to given problem
$K . E . = (75 / 100) (K . E .)_{max}$
$â´ \frac{1}{2} m {Ï}^{2} A^{2} \cos^{2} â¡ Ï t = (\frac{75}{100}) (\frac{1}{2} m {Ï}^{2} A^{2}) or \cos â¡ Ï t = Â\pm \sqrt{3} / 2 or Ï t = \frac{Ï}{6} or \frac{2 Ï}{T} Ã t = \frac{Ï}{6} or t = \frac{T}{12} = \frac{1}{6} s$

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