Question:

Two sources $A$ and $B$ are sending notes of frequency $680 \,Hz$. A listener moves from $A$ and $B$ with-a constant velocity $u$. If the speed of sound in air is $340 \, ms^{-1}$ what must be the value of u so that he hears $10$ beats per second?

Updated On: Jun 3, 2023
  • $2.0 \, ms^{-1}$
  • $2.5 \, ms^{-1}$
  • $3.0 \, ms^{-1}$
  • $3.5 \, ms^{-1}$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Listener go from $A \to B$ with velocity $(u)$
let the apparent frequency of sound from source $A$ by listener
$n' = n \frac{v-v_{0}}{v+v_{s}}$
Or $n' = 680 \frac{340 -u}{340 + 0} $
The apparent frequency of sound from source $B$ by listener
$ n" = n \frac{v+v_{0}}{v-v_{s} } = 680 \frac{340+u}{340-0}$
But listener hear $10$ beats per second.
Or $680 \frac{340 + u}{340} - 680 \frac{340 - u }{340} = 10$
Or $ 2340 + u - 340 + u = 10 $
$ u =2.5 \,ms^{-1}$
Was this answer helpful?
1
0

Top Questions on doppler effect

View More Questions

Concepts Used:

Doppler Effect

The Doppler effect is a phenomenon caused by a moving wave source that causes an apparent upward shift in frequency for observers who are approaching the source and a visible downward change in frequency for observers who are retreating from the source. It's crucial to note that the impact isn't caused by a change in the source's frequency.

 

 

 

 

 

 

 

 

 

 

The Doppler effect may be seen in any wave type, including water waves, sound waves, and light waves. We are most familiar with the Doppler effect because of our encounters with sound waves