JEE Main · 2006 · 30 Apr 2006 · Offline · Elasticity & SHM

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

(A) $\frac{1}{12} s$
(B) $\frac{1}{6} s$ ✓ Correct
(C) $\frac{1}{4} s$
(D) $\frac{1}{3} s$

Answer: (B)

Solution

Starting from the origin (the mean position) the displacement is

x = A sin ω t

. The kinetic energy is

75%

of the total when the potential energy is the other

25%

, that is

\frac{1}{2} k x^{2} = \frac{1}{4} (\frac{1}{2} k A^{2})

, which gives

x = \frac{A}{2}

. Hence

sin ω t = \frac{1}{2}

, so

ω t = \frac{π}{6}

. With

ω = \frac{2 π}{T} = \frac{2 π}{2} = π rad/s

, the time is

t = \frac{π /6}{π} = \frac{1}{6} s

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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