Question

A particle executing simple harmonic motion has a period of $6s$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

• A. $\frac{1}{4}S$
• B. $\frac{3}{4}S$
• C. $\frac{1}{2}S$
• D. $\frac{3}{2}S$

Answer 1

Answer: (C)

$\frac{1}{2}S$

when started from 'o'

$\frac{A}{2}=Asin\omega t$

$\omega t=\frac{\pi }{6}$

$t=\frac{T}{12}$

when started from A

$\frac{A}{2}=Acos\omega t$

$\omega t=\frac{\pi }{3}$

$\frac{2\pi t}{T}=\frac{\pi }{3}$

$t=\frac{T}{6}$

which is half the first therfore 1/2 second is the time taken