Question

A particle executes a simple harmonic motion of time period $T$ . Find the time taken by the particle to go directly from its mean position to half the amplitude

• A. $T/2$
• B. $T/4$
• C. $T/8$
• D. $T/12$

Answer 1

Answer: (D)

$T/12$

Answer 2

$y=a \sin \omega t=\frac{a \sin 2 \pi}{T} t$ At the mean position $y=\frac{a}{2}$ $\frac{a}{2}=\frac{a \sin 2 \pi t}{T}$ $t=\frac{T}{12}$