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Two simple harmonic motions are represented by the equations y_{1}= 0.1\sin \left ( 100\pi t+\frac{\pi }{3} \right )\: and \: y_{2}= 0.1\cos \pi t.

The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

  • Option 1)

    -\pi /3

  • Option 2)

    \pi /6

  • Option 3)

    -\pi /6

  • Option 4)

    \pi /3

 

Answers (1)

As we learnt in

Phase -

The quantity \phi = wt+\delta is called the phase . It determines the status of the particle in simple harmonic motion.

 

 

- wherein

e.g. 

x= A\sin \left ( wt +\delta \right )\rightarrow phase

 

 

 

v_{1}= \frac{d}{dt}\left ( y_{1} \right )= \left ( 0.1\times 100\pi \right )\cos \left ( 100\pi t+\frac{\pi }{3} \right )

v_{2}= \frac{d}{dt}\left ( y_{2} \right )=\left ( -0.1\times \pi \right )\sin \pi t

= \left ( 0.1\times \pi \right )\cos \left ( \pi t+\frac{\pi }{2} \right )

\therefore \: \: \: \Delta \phi = \frac{\pi }{3}-\frac{\pi }{2}= -\frac{\pi }{6}

Correct option is 3.


Option 1)

-\pi /3

This is an incorrect option.

Option 2)

\pi /6

This is an incorrect option.

Option 3)

-\pi /6

This is the correct option.

Option 4)

\pi /3

This is an incorrect option.

Posted by

Vakul

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