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Engineering

Try this! - Oscillations and Waves - JEE Main-2

The phase difference between two points seperated by 0.8 m in  wave of frequency is 120 Hz is \frac{\pi}{}2. The velocity of the wave is 

  • Option 1)

    720 m/s

  • Option 2)

    384 m/s

  • Option 3)

    250 m/s

  • Option 4)

    1 m/s

 
Answers (1)
88 Views
A Avinash

Phase difference = \frac{2 \pi}{\lambda} X path difference \Rightarrow \frac{\pi}{2} = \frac{2 \pi}{\lambda} X 0.8 \Rightarrow \lambda = 4 X 0.8 = 3.2 m

velocity v = n \lambda = 120 X 3.2 = 384 m/s

 

 

In phase and out of phase -

\Delta x= n\lambda \: \: \: or\: \: \Delta \phi = 2n\pi

i.e. point seprated by distance n\lambda are in same phase.

\Delta x= \frac{m\lambda }{2}\left ( m= 1,3,5,7...... \right )\\\ or \: \Delta \phi = m\pi

Point separated by \frac{m\lambda }{2} are out of phase.

-

 

 

 


Option 1)

720 m/s

This is incorrect.

Option 2)

384 m/s

This is correct.

Option 3)

250 m/s

This is incorrect.

Option 4)

1 m/s

This is incorrect.

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