If $\lambda_1 = 100\text{ cm}, \lambda_2 = 90\text{ cm}$ and velocity of $\text{sound} = 396\text{ ms}^{-1}$…

If $\lambda_1 = 100\text{ cm}, \lambda_2 = 90\text{ cm}$ and velocity of $\text{sound} = 396\text{ ms}^{-1}$, the number of beats (in Hz) are
  1. 4
  2. 2
  3. 3
  4. 44

Solution

Fundamental frequency of closed pipe, $f_1 = \frac{v}{4l} = 80\text{ Hz}$. Next is $f_2 = 3f_1 = 240\text{ Hz} \rightarrow \text{first overtone}$ and $f_3 = 5f_1 = 400\text{ Hz} \rightarrow \text{second overtone}$.

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