# Thread: [SOLVED] Time elapsed (SHM)?

1. ## [SOLVED] Time elapsed (SHM)?

A particle executes SHM with period T about a point O. It passes through a point P with velocity V along OP. Show that the time that elapses when it again comes to P is given by: $t = \frac{T}{\pi}\cdot \arctan \left(\frac{TV}{2\pi\cdot OP}\right)$

2. Here's how to do it:

Spoiler:

It is well known that the solution for SHM with ( $x=0$ when $t=0$) is:

$x = A \sin (\omega t)$

with

$\dot{x} = A \omega \cos (\omega t)$

where $A$ is the amplitude of oscillations and $\omega = \frac{2 \pi}{T}$.

Now in the first instance when the particle is at $P$ let the time be $t_0$ and let $d = |\vec{\mathbf{OP}}|$ so then

$d = A \sin (\omega t_0 )$ and $V = A \omega \cos (\omega t_0)$

which after dividing gives us:

$\boxed{\frac{\omega d}{V} = \tan (\omega t_0 )}$ .

Similarly in the second instance when the particle is at $P$ let the time be $t_1$ and so then

$d = A \sin (\omega t_1 )$ and $-V = A \omega \cos (\omega t_1)$

which after dividing gives us:

$\boxed{-\frac{\omega d}{V} = \tan (\omega t_1 )}$. (Note this is in 2nd quadrant).

Thus (taking into account the quadrants) we have:

$\omega (t_1 - t_0 ) = \arctan \left( \frac{\omega d}{-V} \right) - \arctan \left( \frac{\omega d}{V} \right)$
$= \pi - 2 \arctan \left( \frac{\omega d}{V} \right)$

so

$\frac{\omega d}{V} = \tan \left( \frac{\pi}{2} - \frac{\omega}{2} (t_1 - t_0 ) \right)$

$= \cot \left( \frac{\omega}{2} (t_1 - t_0 ) \right)$

$= \cot \left( \frac{\pi}{T} (t_1 - t_0 ) \right)$

$\Leftrightarrow \tan \left( \frac{\pi}{T} (t_1 - t_0 ) \right) = \frac{V}{\omega d}$

Hence

${\color[rgb]{0,0,1} \boxed{t = t_1 - t_0 = \frac{T}{\pi} \arctan \left( \frac{TV}{2 \pi d} \right)}}$ .