Two particles are executing simple harmonic motion of the same amplitude A and frequency \omega along the x-axis. Their mean position is separated by distance X_{0}(X_{0}> A). If the maximum separation between them is (X_{0}+A), the phase difference between their motion is

  • Option 1)

    \frac{\pi }{2}

  • Option 2)

    \frac{\pi }{3}

  • Option 3)

    \frac{\pi }{4}

  • Option 4)

    \frac{\pi }{6}

 

Answers (1)

As we learnt in

Simple harmonic as projection of circular motion -

- wherein

x= A\cos \omega t

y= A\sin wt

 

 x_{1}=x_{0}cos\omega t

x_{2}=x_{0}cos(\omega t+\phi)

The crossing happens when x_{1}=\frac {x_{0}}{2}

\frac {x_{0}}{2}=x_{0}cos \omega t=cos\omega t=\frac{1}{2}

cos\omega t=cos\frac{\pi}{3}

\omega t=\frac{\pi}{3}

Correct option is 2.

 


Option 1)

\frac{\pi }{2}

This is an incorrect option.

Option 2)

\frac{\pi }{3}

This is the corect option.

Option 3)

\frac{\pi }{4}

This is an incorrect option.

Option 4)

\frac{\pi }{6}

This is an incorrect option.

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