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 \(340\,m\,s^{-1}\), what must be the value of \(u\) so that he hears \(10\) beats per second?

• A. \(20\,m\,s^{-1}\)
• B. \(2.5\,m\,s^{-1}\)
• C. \(3.0\,m\,s^{-1}\)
• D. \(3.5\,m\,s^{-1}\)

Hint:

For beats with Doppler effect, use \(f_b = f\left(\frac{2u}{v}\right)\) when listener moves towards one source and away from the other.

Answer 1

The Correct Option is B

Step 1: Understand beat frequency.
Beat frequency is the difference between the two frequencies heard by the listener.
\[ f_b = |f'_A - f'_B| \]
Here, both sources emit same frequency \(f = 680\,Hz\).
Step 2: Apply Doppler effect.
Listener is moving away from both sources, but relative direction is different:
- From one source, listener is moving away \(\Rightarrow\) frequency decreases.
- From the other source (if opposite direction considered), listener is moving towards \(\Rightarrow\) frequency increases.
So:
\[ f'_1 = f\left(\frac{v-u}{v}\right) ,\quad f'_2 = f\left(\frac{v+u}{v}\right) \]
Step 3: Find beat frequency.
\[ f_b = f'_2 - f'_1 = f\left(\frac{v+u}{v}\right) - f\left(\frac{v-u}{v}\right) \]
\[ f_b = f\left(\frac{2u}{v}\right) \]
Step 4: Substitute values.
Given: \(f_b = 10\,Hz\), \(f = 680\,Hz\), \(v = 340\,m\,s^{-1}\).
\[ 10 = 680\left(\frac{2u}{340}\right) \]
\[ 10 = 680\left(\frac{u}{170}\right) \Rightarrow 10 = 4u \Rightarrow u = 2.5\,m\,s^{-1} \]
Final Answer: \[ \boxed{2.5\,m\,s^{-1}} \]