Question

The position of a particle moving along X-axis varies with time as x = ut + Asinwt . At t = 0, x = 0 . Between t = 0 and t = t0 , average velocity of particle is u . Then values of t0 can be

Answer 1

Given:

Position of particle moving along x axis varies as follows ;

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

At t = 0, x = 0 and the average velocity in between the time t = 0 and t = t0 is u.

To find:

Possible values of $t_{0}$

Calculation:

Average Velocity is defined as the ratio of total displacement to the total time taken to cover that specified displacement.

We can represent total displacement as a difference between the position of the object .

$\therefore \: avg. \: v = \dfrac{total \: displacement}{total \: time}$

$= > \: u = \dfrac{x2 - x1}{t2 - t1}$

$= > \: u = \dfrac{ \bigg \{ut_{0} + A \sin( \omega t_{0}) \bigg \} - 0}{t_{0} - 0}$

$= > \: u = \dfrac{ ut_{0} + A \sin( \omega t_{0})}{t_{0} }$

$= > \: u = u + \dfrac{ A \sin( \omega t_{0})}{t_{0} }$

$= > \: \dfrac{ A \sin( \omega t_{0})}{t_{0} } = 0$

Since denominator can't be zero , we can say that :

$\therefore \: A \sin( \omega t_{0}) = 0$

$= > \: \sin( \omega t_{0}) = 0$

Considering general solutions , we get :

$= > \: \sin( \omega t_{0}) = \sin(n\pi) \: \: \: \: \: \: \{n \in I \}$

$= > \: \omega t_{0} = n\pi$

$= > \: t_{0} = \dfrac{ n\pi}{ \omega }$

Putting values of n as 1 , 2 and 4 , we get

$= > \: t_{0} = \dfrac{ \pi}{ \omega } \: or \: \dfrac{ 2\pi}{ \omega } \: or \: \dfrac{ 4\pi}{ \omega }$

So final answer is :

$\boxed{ \red{ \large{ \bold{ t_{0} = \dfrac{ \pi}{ \omega } \: or \: \dfrac{ 2\pi}{ \omega } \: or \: \dfrac{ 4\pi}{ \omega } }}}}$