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  • Stuck here, help me understand: - Oscillations and Waves - JEE Main-7

 

Two waves  having equations  x_{1} = a\sin(\omega t + \phi_{2} ),  x_{2} = a\sin(\omega t + \phi_{2}). If in the resultant wave the frequency and amplitude remain equal tothose superimposing waves. Then phase difference between them is 

  • Option 1)

    \frac{\pi}{6}

  • Option 2)

    \frac{2\pi}{3}

  • Option 3)

    \frac{\pi}{4}

  • Option 4)

    \frac{\pi}{3}

 

Answers (1)

best_answer

Superposition of waves does not alter the freaquency of resultant  wave and resultant amplitude

^{2}= a^{2}+ a^{2}+ 2a^{2} \cos \Phi = 2a^{2}(1+\cos \Phi )

\cos \Phi = -1/2 = \cos 2\pi /3

\therefore 2\pi /3

 

Resultant Intensity -

I=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}.\cos \phi

 

- wherein

\phi = phase\: dif\! ference

 

 


Option 1)

\frac{\pi}{6}

This is incorrect

Option 2)

\frac{2\pi}{3}

This is correct

Option 3)

\frac{\pi}{4}

This is incorrect

Option 4)

\frac{\pi}{3}

This is incorrect

Posted by

Plabita

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