Question

The fundamental frequency of a closed organ pipe is $220Hz$ . The second overtone of this pipe has the same wavelength as the third harmonic of an open pipe. Take speed of sound in air $345m{s^{ - 1}}$ . The length of this pipe is $470 \times {10^{ - x}}m$. Find $x$.

Answer 1

Here, the organ pipe mentioned is closed and its fundamental frequency is given. Also it has been given that the wavelengths of the second overtone of the closed pipe and the third overtone of the open pipe are equal or the same. We know that the fundamental frequency for open organ pipe is given by the formula: ${f_n} = n\dfrac{v}{{2L}}$ , similarly for closed organ pipe the fundamental frequency is given by: ${f_n} = n\dfrac{v}{{4L}}$
Overtones of closed pipe and open pipe are given by the formula as: ${\lambda _n} = \dfrac{v}{{{f_n}}}$
Using these formulas calculate the wavelengths of required overtones and then find the required answer.

Complete answer:
According to the given information we have:
Fundamental frequency of closed organ pipe: $f = 220Hz$
Speed of sound, $v = 345m{s^{ - 1}}$
Let $L$ be the length of a closed organ pipe and $L'$ be the length of an open pipe.
Wavelength of second overtone in closed pipe,
$\lambda = \dfrac{{4L}}{5}$ …. (since, ${\lambda _n} = \dfrac{v}{{{f_n}}}$ )
Wavelength of third harmonic in open pipe,
$\lambda ' = \dfrac{{2L'}}{3}$
And also it is given that $\lambda = \lambda '$ ,
Thus,
$\dfrac{{4L}}{5} = \dfrac{{2L'}}{3}$
$\therefore L' = \dfrac{{12L}}{{10}}$
Now, the fundamental frequency of closed organ pipe is:
$f = 220Hz$ , we have
$\Rightarrow 220 = \dfrac{v}{{4L}}$
$\Rightarrow L = \dfrac{{345}}{{4 \times 220}} = 0.392m$
$\therefore L' = \dfrac{{12 \times 0.392}}{{10}} = 0.470m$
i.e. $L' = 470 \times {10^{ - 3}}m$
On comparing $470 \times {10^{ - x}}m$ we get the value of $x$ as $3$ .
Hence, the answer is $3$ .

Note:
We have to use the formulas carefully and must understand the meaning of overtone and harmonic i.e. second harmonic is the first overtone. This concept is used to calculate the wavelength of the open organ pipe. We have followed the procedure and calculated all the values carefully.