Question

A particle performs shm on x axis with amplitude a and time period t. The time taken by the particle to travel a distance a/5 starting from rest is ??

Answer 1

Answer: The time taken by the particle is $t = \dfrac{T}{2\pi}\ cos^{-1}(\dfrac{1}{5})$

Explanation:

Given that,

Amplitude = a

Time period = t

Distance $y = \dfrac{a}{5}$

The general equation of SHM

$y = a\ cos \omega t$....(I)

Put the value of y in equation (I)

$\dfrac{a}{5} = a\ cos \omega t$

$\omega t = cose^{-1}(\dfrac{1}{5})$

$t = \dfrac{1}{\omega}\ cos^{-1}(\dfrac{1}{5})$

We know that,

$\omega = \dfrac{2\pi}{T}$

Now, the time taken by the particle

$t = \dfrac{T}{2\pi}\ cos^{-1}(\dfrac{1}{5})$

Hence, this is the required solution.