The frequency of open organ pipe is | Physics - Standing Waves in Pipes | SolveeIt

The frequency of open organ pipe is

$\frac{v}{2l+0.4d}$

$\frac{v}{2l+0.6d}$

$\frac{v}{2l+0.8d}$

$\frac{v}{2l+1.2d}$

Answer

$\frac{v}{2l+0.6d}$

Explanation

For an open organ pipe, the effective length is given by:

$L_{\text{eff}} = l + 0.3d \quad \text{(for one end)}$

Since both ends are open, the total effective length is:

$2l + 2(0.3d) = 2l + 0.6d$

Thus, the frequency of the fundamental mode is:

$f = \frac{v}{2l + 0.6d}$

Question: The frequency of open organ pipe is...