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Two simple harmonic motions are represented by the equations y = 0.1\sin(100\pi t + \frac{\pi}{3}) and y = 0.1\cos(\pi t) . The phase difference of the velocity of the particle 1 with respect to the velocity of particle 2 is 

  • Option 1)

    \frac{-\pi}{3}

  • Option 2)

    \frac{\pi}{6}

  • Option 3)

    \frac{-\pi}{6}

  • Option 4)

    \frac{\pi}{3}

 

Answers (1)

best_answer

\\*v_{1} = \frac{dy_{1}}{dt} = 0.1\times 100\pi \cos(100\pi t + \frac{\pi}{3}) \\*v_{1} = \frac{dy_{1}}{dt} = 0.1\times \pi \sin(\pi t) = 0.1\pi \cos (\pi t + \frac{\pi}{2})

Phase Difference of velocity of first particle with respect to the velocity of 2nd particle at t = 0 is 

\Delta\phi = \phi_{1} - \phi_{2} = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6}

 

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

 

 

 

 


Option 1)

\frac{-\pi}{3}

This is incorrect.

Option 2)

\frac{\pi}{6}

This is incorrect.

Option 3)

\frac{-\pi}{6}

This is correct.

Option 4)

\frac{\pi}{3}

This is incorrect.

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