A pipe closed at one end and open at the other end resonates with a sound of frequency $135\text{ Hz}$ and…

A pipe closed at one end and open at the other end resonates with a sound of frequency $135\text{ Hz}$ and also with $165\text{ Hz}$, but not at any other frequency intermediate between these two. Then, the frequency of the fundamental note of the pipe is [EAMCET 2013]
  1. $15\text{ Hz}$
  2. $60\text{ Hz}$
  3. $7.5\text{ Hz}$
  4. $30\text{ Hz}$

Solution

The frequency of the fundamental note of the pipe is $n_2 - n_1 = 2\left(\frac{v}{4l}\right) \Rightarrow \frac{v}{4l} = \frac{n_2 - n_1}{2}$ Given, $n_1 = 135\text{ Hz}$ and $n_2 = 165\text{ Hz}$ Now, $\frac{v}{4l} = \frac{165 - 135}{2} = \frac{30}{2} = 15\text{ Hz}$

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