Two sources $A$ and $B$ are sounding notes of frequency $680\text{ Hz}$. A listener moves from $A$ to $B$…

Two sources $A$ and $B$ are sounding notes of frequency $680\text{ Hz}$. A listener moves from $A$ to $B$ with a constant velocity $u$. If the speed of sound is $340\text{ ms}^{-1}$, what must be the value of $u$, so that he hears 10 beats per second?
  1. $2.0\text{ ms}^{-1}$
  2. $2.5\text{ ms}^{-1}$
  3. $3.0\text{ ms}^{-1}$
  4. $3.5\text{ ms}^{-1}$

Solution

As, per question, $f_1 - f_2 = 10$ $\cdot\!\cdot\!\cdot\, (680) \left(\frac{340 + u}{340}\right) - (680) \left(\frac{340 - u}{340}\right) = 10 \Rightarrow u = 2.5\text{ ms}^{-1}$

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