Question

The time taken by a particle executing simple harmonic motion of time period T to move from the mean position to half the maximum displacement is

• A. $\frac{T}{2}$
• B. $\frac{T}{4}$
• C. $\frac{T}{8}$
• D. $\frac{T}{12}$

Answer 1

The correct option is D $\frac{T}{12}$

Let the displacement of the particle be given by x = A sin $\omega$ t = A sin $\left(\frac{2\pi t}{T}\right)$ i.e., when x = 0, $t_0=0.$
When x = A/2, the value of t is given by $\frac{A}{2}=Asin\frac{2\pi t_1}{T};sin\frac{2\pi t_1}{T}=\frac{1}{2}$
$\frac{2\pi t_1}{T}=\frac{\pi }{6}or{t}_1=\frac{T}{12};t_1-t_0=\frac{T}{12}-0=\frac{T}{12}$