Question 299: Waves

Question

An open organ pipe of length ' $l$ ' is sounded together with another open organ pipe of length $\left(l+l_1\right)$ in their fundamental modes. Speed of sound in air is ' $V$ '. The beat frequency heard will be ( $\left.l_1< < l\right)$

Accepted Answer

For an open organ pipe, the fundamental frequency is given by:
$ f = \frac{V}{2l} $
where $ V $ is the speed of sound in air, and $ l $ is the length of the organ pipe.
Given two open pipes with lengths $ l $ and $ l + l_1 $ respectively, their fundamental frequencies will be:
$ f_1 = \frac{V}{2l} $
$ f_2 = \frac{V}{2(l + l_1)} $
The beat frequency, which is the difference between these two frequencies, is:
$ f_b = f_1 - f_2 = \frac{V}{2l} - \frac{V}{2(l + l_1)} $
Simplifying, the equation becomes:
$ f_b = V \left[ \frac{2(l + l_1) - 2l}{4l(l + l_1)} \right] = V \frac{2l_1}{4l(l + l_1)} $
Further simplification gives:
$ f_b = \frac{V l_1}{2l(l + l_1)} $
Assuming $ l_1 \ll l $, we can approximate the denominator by taking $ (l + l_1) \approx l $:
$ f_b = \frac{V l_1}{2l^2} $

Answer

$\frac{\mathrm{V} l_1^2}{2 l}$

Answer

$\frac{\mathrm{V} l_1}{2 l^2}$

Answer

$\frac{\mathrm{V} l_1}{4 l^2}$

Answer

$\frac{\mathrm{v} l^2}{2 l_1}$

Waves

An open organ pipe of length ' l ' is sounded together with another open organ pipe of length \left(l+l_1\right) in their fundamental modes. Speed of sound in air is ' V '. The beat frequency heard will be ( \left.l_1< < l\right)

Choose the correct answer: