Question

The 4 ^ (th) overtone of a closed organ pipe is same as that of 3 ^ (rd) overtone of an open pipe. The ratio of the length of the closed pipe to the length of organ pipe

Answer 1

9/8

Answer 2

To solve this problem, we need to use the formulas for the frequencies of overtones in closed and open organ pipes.

1. Frequencies in a Closed Organ Pipe:

For a closed organ pipe, only odd harmonics are present. The general formula for the frequency of the $n^{th}$ harmonic is:

$f_n = \frac{nv}{4L_c}$

where $n = 1, 3, 5, \dots$

The fundamental frequency is the 1st harmonic ($n=1$).
The 1st overtone is the 3rd harmonic ($n=3$).
The 2nd overtone is the 5th harmonic ($n=5$).
In general, the $k^{th}$ overtone corresponds to the $(2k+1)^{th}$ harmonic.

For the 4th overtone of a closed organ pipe, $k=4$. So, the harmonic number is $(2 \times 4 + 1) = 9$.
The frequency of the 4th overtone of the closed pipe is:

$f_{c, 4th\_overtone} = \frac{9v}{4L_c}$

where $L_c$ is the length of the closed pipe and $v$ is the speed of sound.

2. Frequencies in an Open Organ Pipe:

For an open organ pipe, all harmonics are present. The general formula for the frequency of the $n^{th}$ harmonic is:

$f_n = \frac{nv}{2L_o}$

where $n = 1, 2, 3, \dots$

The fundamental frequency is the 1st harmonic ($n=1$).
The 1st overtone is the 2nd harmonic ($n=2$).
The 2nd overtone is the 3rd harmonic ($n=3$).
In general, the $k^{th}$ overtone corresponds to the $(k+1)^{th}$ harmonic.

For the 3rd overtone of an open organ pipe, $k=3$. So, the harmonic number is $(3+1) = 4$.
The frequency of the 3rd overtone of the open pipe is:

$f_{o, 3rd\_overtone} = \frac{4v}{2L_o} = \frac{2v}{L_o}$

where $L_o$ is the length of the open pipe.

3. Equating the Frequencies:

According to the problem statement, the 4th overtone of the closed pipe is the same as the 3rd overtone of the open pipe:

$f_{c, 4th\_overtone} = f_{o, 3rd\_overtone}$

$\frac{9v}{4L_c} = \frac{2v}{L_o}$

4. Solving for the Ratio of Lengths:

Cancel $v$ from both sides of the equation:

$\frac{9}{4L_c} = \frac{2}{L_o}$

To find the ratio $\frac{L_c}{L_o}$, rearrange the equation:

$9L_o = 2 \times 4L_c$

$9L_o = 8L_c$

Now, divide both sides by $8L_o$:

$\frac{L_c}{L_o} = \frac{9}{8}$

The ratio of the length of the closed pipe to the length of the open pipe is $9/8$.