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Equation of motion in the same direction are given by y_{1} = 2a\sin(\omega t -kx) and y = 2a\sin(\omega t -kx -\theta). The amplitude of the medium particle will be

  • Option 1)

    2a\cos\theta

  • Option 2)

    \sqrt{2}a\cos\theta

  • Option 3)

    4a\cos\frac{\theta}{2}

  • Option 4)

    \sqrt{2}a\cos\frac{\theta}{2}

 

Answers (1)

best_answer

Resultant amplitude 

A_{R}= 2a \cos \left ( \frac{\Theta }{2} \right )= 2\times 2a \cos \left ( \frac{\Theta }{2} \right ) = 4a \cos \left ( \frac{\Theta }{2} \right )

 

Resultant Intensity -

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

 

- wherein

\phi = phase\: dif\! ference

 

 


Option 1)

2a\cos\theta

This is incorrect

Option 2)

\sqrt{2}a\cos\theta

This is incorrect

Option 3)

4a\cos\frac{\theta}{2}

This is correct

Option 4)

\sqrt{2}a\cos\frac{\theta}{2}

This is incorrect

Posted by

divya.saini

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