Q

# 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

88 Views

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.

Exams
Articles
Questions