The $3^{\text {rd }}$ overtone of a closed organ pipe is in unison with $t^{\text {th}}$ harmonic of an open…
The $3^{\text {rd }}$ overtone of a closed organ pipe is in unison with $t^{\text {th}}$ harmonic of an open orhan pipe. The ratio of the length of the closed pipe to length of open pipe is
$\frac{7}{8}$
$\frac{4}{3}$
$\frac{6}{5}$
$\frac{7}{9}$
Solution
To solve the problem of finding the ratio of the lengths of a closed organ pipe to an open organ pipe when their frequencies are in unison, we can follow these steps:
Step 1: Understand the frequencies of the pipes
- For a closed organ pipe, the frequencies are given by:
\(f_n=\frac{(2 n-1) V}{4 L_1}\)
where \(n\) is the harmonic number (1st, 2nd, 3rd, ...), \(V\) is the speed of sound, and \(L_1\) is the length of the closed organ pipe.
- For an open organ pipe, the frequencies are given by:
\(f_{\mathrm{n}}=\frac{\mathrm{nV}}{2 L_2}\)
where \(L_2\) is the length of the open organ pipe.
Step 2: Identify the specific overtone and harmonic
-The third overtone of a closed organ pipe corresponds to the 7th harmonic:
\(f_{3 \text { overtone }}=f_7=\frac{7 V}{4 L_1}\)
- The fourth harmonic of an open organ pipe is:
\(f_{4 \text { harmonic }}=f_4=\frac{4 V}{2 L_2}=\frac{2 V}{L_2}\)
Step 3: Set the frequencies equal
Since the third overtone of the closed organ pipe is in unison with the fourth harmonic of the open organ pipe, we can set their frequencies equal:
\(\frac{7 V}{4 L_1}=\frac{2 V}{L_2}\)
Step 4: Simplify the equation
We can cancel \(V\) from both sides (assuming \(V \neq 0\) ):
\(\frac{7}{4 L_1}=\frac{2}{L_2}\)
Step 5: Cross-multiply to find the ratio of lengths
Cross-multiplying gives:
\(7 L_2=8 L_1\)
Step 6: Solve for the ratio of lengths
Rearranging the equation gives us:
\(\frac{L_1}{L_2}=\frac{7}{8}\)
Final Answer
The ratio of the lengths of the closed organ pipe to the open organ pipe is:
\(\frac{L_1}{L_2}=\frac{7}{8}\)
.