Question

A particle executes SHM along a straight line so that its period is 12 s. The time it takes in traversing a distance equal to half its amplitude from its equilibrium position is-

• A. 6 s
• B. 4 s
• C. 2 s
• D. 1s

Answer 1

Answer: (D)

1s

Answer 2

E = a sin ωt ⇒ $\frac{a}{2} = a\sin\omega t$ or $\frac{1}{2} = \sin\omega t$

or $\sin\frac{\pi}{6} = \sin\frac{2\pi}{12}t$ ⇒ t = 1s