Step 1: Understanding the Concept:
Organ pipes produce standing waves. The frequencies of these standing waves depend on whether the pipe is open at both ends or closed at one end.
A closed pipe supports only odd harmonics, while an open pipe supports all integer harmonics.
Step 2: Key Formulas or Approach:
For a closed pipe of length \( L_c \), the allowed frequencies are \( f_n = (2n + 1)\frac{v}{4L_c} \), where \( n = 0, 1, 2, \dots \) represents the overtone number.
For an open pipe of length \( L_o \), the allowed frequencies are \( f'_m = (m + 1)\frac{v}{2L_o} \), where \( m = 0, 1, 2, \dots \) represents the overtone number.
Step 3: Detailed Explanation:
Let's find the frequency of the fourth overtone of the closed pipe.
Here, \( n = 4 \).
Frequency \( f_c = (2(4) + 1)\frac{v}{4L_c} = \frac{9v}{4L_c} \).
This corresponds to the 9th harmonic.
Now, let's find the frequency of the fifth overtone of the open pipe.
Here, \( m = 5 \).
Frequency \( f_o = (5 + 1)\frac{v}{2L_o} = \frac{6v}{2L_o} = \frac{3v}{L_o} \).
This corresponds to the 6th harmonic.
The problem states that these two frequencies are in unison, meaning they are equal.
Equating the two frequencies:
\[ \frac{9v}{4L_c} = \frac{3v}{L_o} \]
Cancel the speed of sound \( v \) from both sides:
\[ \frac{9}{4L_c} = \frac{3}{L_o} \]
Rearrange to find the ratio \( \frac{L_c}{L_o} \):
\[ \frac{L_c}{L_o} = \frac{9}{4 \times 3} \]
\[ \frac{L_c}{L_o} = \frac{9}{12} \]
Simplify the fraction:
\[ \frac{L_c}{L_o} = \frac{3}{4} \]
The ratio is \( 3 : 4 \).
Step 4: Final Answer:
The ratio of length of closed pipe to that of open pipe is \( 3 : 4 \).