Question

The time taken by a particle executing SHM of period T to move from the mean position to half the maximum displacement is:
1) T/2
2)T/4
3)T/8
4)T/12

Answer 1

(4) T/12.

The phase difference between the mean position, and half the maximum displacement (while moving away from the mean position) is $\frac{\pi}{6}$.

We know that T corresponds to the complete cycle, that is, $2\pi$ phase.

So, the required time is: $\frac{T}{2\pi} * \frac{\pi}{6} = \frac{T}{12}$.

Answer 2

Given :

Period of SHM is T .

To Find :

Time taken by a particle to move from the mean position to half the maximum displacement .

Solution :

We know , displacement in SHM is given by :

$x=A\ sin\ \omega t$

Here , $\omega=\dfrac{2\pi}{T}$ .

Let , at  half the maximum displacement  time be t .

Therefore ,

$\dfrac{A}{2}=A\ sin\ \omega t\\\\sin\ \omega t=\dfrac{1}{2}$

Therefore ,

$\omega t=\dfrac{\pi}{6}\\\\\dfrac{2\pi}{T}\times t=\dfrac{\pi}{6}\\\\t=\dfrac{T}{12}$

Learn More :

Simple Harmonic Motion

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