Physics - Sound Wave Question with Solution | TestHub

Length of the organ pipe is same in both the cases. Fundamental frequency of open pipe is${\text{f}}_1=\frac{\text{v}}{2\mathit{\text{l}}}$Frequency of third harmonic of closed pipe will be.

$f_2=3\left(\frac{v}{4l}\right)$

Given that,$f_2=f_1+100$

or$f_2-f_1=100$

or$\frac{3}{4}\left(\frac{v}{l}\right)-\left(\frac{1}{2}\right) \left(\frac{v}{l}\right)=100\Rightarrow \frac{v}{4l}=100$

$\therefore$$\frac{v}{2l}\text{ or }{f}_1=200\text{ Hz}$$\text{50}\text{.}\text{n}\text{Hz}\Rightarrow \text{n = 4}$

Therefore, fundamental frequency of the open pipe is 200 Hz. and$\text{n = 4}$