Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two particles are performing simple harmonic motion in a straight line about the same equilibrium point.  The amplitude and time period for both particles are same and equal to A and<

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point.  The amplitude and time period for both particles are same and equal to A and T, respectively.  At time t = 0 one particle has displacement A while the other one has displacement  \frac{-A}{2}    and they are moving towards each other.  If they cross each other at time t, then t is :

 

Option: 1

\frac{T}{6}


Option: 2

\frac{5T}{6}


Option: 3

\frac{T}{3}


Option: 4

\frac{T}{4}


Answers (1)

best_answer

As we learned in

Simple harmonic as a projection of circular motion -

- wherein

x= A\cos \omega t

y= A\sin wt

Angle covered to meet \theta=60=\frac{\pi}{3}rad

If any cross each other at time t then 

t=\frac{\theta}{\omega}=\frac{\pi}{3}\times\frac{T}{2\pi}

\therefore\ \;t=\frac{T}{6}

The correct option is 1.

 

Posted by

manish

View full answer

Similar Questions

Latest Question