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Equation of a progressive wave is given by y = 0.2\cos\pi(0.04 +0.02x -\frac{\pi}{6}). The distance is expressed in cm and time is second. What will the minimum distance between two particles have the phase difference of  \frac{\pi}{}2

  • Option 1)

    4 cm

  • Option 2)

    8 cm 

  • Option 3)

    25 cm

  • Option 4)

    12.5 cm

 

Answers (1)

Comparing with y = a cos (\omega t + kx - \phi ) ,

we get k = \frac{2 \pi}{\lambda} = 0.02 \pi \Rightarrow \lambda = 100 cm

Also it is given that phase difference between particles

\Delta \phi = \frac{\pi}{2}.

Hence path difference between them \Delta = \frac{\lambda}{2} X \Delta \phi = \frac{\lambda}{2\pi} X \frac{\pi}{2} = \frac{\lambda}{4} = \frac{100}{4} = 25 cm

 

Relation between phase difference and path difference -

Phase difference \left ( \Delta \phi \right )

= \frac{2\pi }{\lambda }\times path \: dif\! \! ference\left ( \Delta x \right )\\\lambda =wave\; length

-

 

 

 


Option 1)

4 cm

This is incorrect.

Option 2)

8 cm 

This is incorrect.

Option 3)

25 cm

This is correct.

Option 4)

12.5 cm

This is incorrect.

Posted by

Vakul

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