The fundamental frequency of a closed pipe is $220\text{ Hz}$. If $\left(\frac{1}{4}\right)$ th of the pipe…
The fundamental frequency of a closed pipe is $220\text{ Hz}$. If $\left(\frac{1}{4}\right)$ th of the pipe is filled with water, then the frequency of the first overtone of the pipe now is [Haryana PMT 2011]
220 Hz
440 Hz
880 Hz
1760 Hz
Solution
Fundamental frequency of closed pipe,
$n = \frac{v}{4l} = 220\text{ Hz}$
$\Rightarrow v = 220 \times 4l$ ...(i)
If $\left(\frac{1}{4}\right)$ th of the pipe is filled with water, then remaining length of air column is $\frac{3l}{4}$.
Now, fundamental frequency $= \frac{v}{4\left(\frac{3l}{4}\right)} = \frac{v}{3l}$
First overtone $= 3 \times \text{Fundamental frequency}$
$= 3 \times \frac{v}{3l} = \frac{3 \times 220 \times 4l}{3l}$ [From Eq. (i)]
$= 880\text{ Hz}$