The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat…
The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the lengths of the pipes (Take, speed of sound in air is v = 330 ms -1)
Solution
Sol. Let $l_1$ and $l_2$ be the lengths of closed and open pipes, respectively.
Fundamental frequency of closed organ pipe is given by
$f_1 = \frac{v}{4 l_1}$
Here, $v$ = speed of sound in air = 330 ms$^{-1}$
But $f_1 = 110$ Hz (Given)
Therefore, $\frac{v}{4 l_1} = 110\ \text{Hz}$
$\Rightarrow\ \; l_1 = \frac{v}{4 \times 110} = \frac{330}{4 \times 110}\ \text{m} = 0.75\ \text{m}$
Frequency of first overtone of closed organ pipe,
$f_3 = 3 f_1 = 3 (110)\ \text{Hz} = 330\ \text{Hz}$
This produces a beat frequency of 2.2 Hz with first overtone of open organ pipe.
Therefore, first overtone frequency of open organ pipe is either
$(330 + 2.2)\ \text{Hz} = 332.2\ \text{Hz}$ or $(330 - 2.2)\ \text{Hz} = 327.8\ \text{Hz}$
If first overtone frequency is 332.2 Hz, then
$2\left(\frac{v}{2 l_2}\right) = 332.2\ \text{Hz}$ or $l_2 = \frac{v}{332.2} = \frac{330}{332.2}\ \text{m} = 0.99\ \text{m}$
and if it is 327.8 Hz, then
$2\left(\frac{v}{2 l_2}\right) = 327.8\ \text{Hz}$ or $l_2 = \frac{v}{327.8}\ \text{m} = \frac{330}{327.8}\ \text{m} = 1.0067\ \text{m}$
Therefore, length of the closed organ pipe is $l_1 = 0.75\ \text{m}$ while length of open pipe is either $l_2 = 0.99\ \text{m}$ or $1.0067\ \text{m}$.
Answer: $l_1 = 0.75\ \text{m},\ l_2 = 0.99\ \text{m}\ \text{or}\ 1.0067\ \text{m}$