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Two waves are represented by x_1=A \sin (\omega t +\frac{\pi}{6}) and x_2=A \cos \omega t, then the phase difference between them is

  • Option 1)

    \frac{\pi}{6}

  • Option 2)

    \frac{\pi}{2}

  • Option 3)

    \frac{\pi}{3}

  • Option 4)

    \pi

 

Answers (1)

best_answer

As we learnt in 

Travelling Wave Equation -

y=A \sin \left ( Kx-\omega t \right )
 

- wherein

K=2\pi /\lambda

\omega = \frac{2\pi }{T}

\lambda =  wave length

T = Time period of oscillation

 

 x_{1}=Asin(\omega t + \frac{\pi}{6})\:\:\:,\:\:\:\:x_{2}=Acos \omega t

                                                     =Asin (\omega t + \frac{\pi}{2})

Phase difference  =(\frac{\pi}{2}-\frac{\pi }{6})=\frac{\pi }{3}

 

 


Option 1)

\frac{\pi}{6}

This option is incorrect.

Option 2)

\frac{\pi}{2}

This option is incorrect.

Option 3)

\frac{\pi}{3}

This option is correct.

Option 4)

\pi

This option is incorrect.

Posted by

Plabita

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