Question

The displacement x of a body in motion is given by x = a sin(ot +0). The time at which the displacement
is maximum is​

Answer 1

Given:

Displacement equation has been provided as follows ;

$x = a \sin( \omega t + \theta)$

To find:

Time at which the displacement of particle will be max ?

Calculation:

$x = a \sin( \omega t + \theta)$

$= > \dfrac{dx}{dt} = a \: \dfrac{d \{ \sin( \omega t + \theta) \} }{dt}$

$= > \dfrac{dx}{dt} = a \omega \cos( \omega t + \theta)$

[ Considering $\theta$ to be a constant ]

For maxima :

$\therefore \dfrac{dx}{dt} = a \omega \cos( \omega t + \theta) = 0$

$= > a \omega \cos( \omega t + \theta) = 0$

$= > \cos( \omega t + \theta) = 0$

Considering general solutions , we get :

$= > \cos( \omega t + \theta) = \cos \{(2n + 1) \dfrac{\pi}{2} \}$

$= > \omega t + \theta = (2n + 1) \dfrac{\pi}{2}$

$= > \omega t = (2n + 1) \dfrac{\pi}{2} - \theta$

$= > t = \bigg \{ \dfrac{ (2n + 1) \dfrac{\pi}{2} - \theta}{ \omega} \bigg \}$

[ Here n $\epsilon$ I ]

So the final answer is :

$\boxed{ \red{ \huge{ \bold{ t = \bigg \{ \dfrac{ (2n + 1) \dfrac{\pi}{2} - \theta}{ \omega} \bigg \}}}}}$