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  • How to solve this problem- - Oscillations and Waves - JEE Main-6

A simple harmonic progressive wave is represented by the equation y = 8\sin(0.1x -2t), where  x\; \&\; y are in cm and t is in second. At any instant the phase difference between two particles seperated by 2  cm is x- direction is 

  • Option 1)

    18\degree

  • Option 2)

    36\degree

  • Option 3)

    54\degree

  • Option 4)

    72\degree

 

Answers (1)

best_answer

y = a sin 2\pi \left [ \frac{t}{T} - \frac{x}{\lambda}\right ] \Rightarrow \lambda = 10 cm

\phi = \frac{2\pi}{\lambda} X path difference \Rightarrow \frac{2\pi}{10} X 2 = \frac{2}{5} X 180^{o} = 72^{o}

 

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)

18\degree

This is incorrect.

Option 2)

36\degree

This is incorrect.

Option 3)

54\degree

This is incorrect.

Option 4)

72\degree

This is correct.

Posted by

prateek

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