Question

The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is \( \frac{5}{x} \). The value of \( x \) is ________.

• A. 2
• B. 3
• C. 4
• D. 1

Hint:

Closed pipes support only odd harmonics, while open pipes support all harmonics.

Answer 1

The Correct Option is A

To determine the value of \( x \), let us analyze the harmonic frequencies for both closed and open organ pipes.

Closed Organ Pipe

The harmonics for a closed organ pipe are given by:

• The fundamental frequency (first harmonic) is \( f_1 = \frac{v}{4L_1} \), where \( L_1 \) is the length of the closed pipe.
• Only odd harmonics are present: \( f_3 = 3f_1, f_5 = 5f_1, \ldots \)

Open Organ Pipe

The harmonics for an open organ pipe are given by:

• The fundamental frequency (first harmonic) is \( f_1 = \frac{v}{2L_2} \), where \( L_2 \) is the length of the open pipe.
• All harmonics are present: \( f_2 = 2f_1, f_3 = 3f_1, \ldots \)

Given Condition

According to the problem, the fifth harmonic of the closed pipe is in unison with the first harmonic of the open pipe.

This can be expressed as:

\(f_5^{(\text{closed})} = f_1^{(\text{open})}\)

Substituting the expressions for these harmonics, we have:

\(\frac{5v}{4L_1} = \frac{v}{2L_2}\)

We can simplify this equation by canceling \( v \) from both sides:

\(\frac{5}{4L_1} = \frac{1}{2L_2}\)

Cross-multiplying gives:

\(10L_2 = 4L_1\)

Simplifying, we find:

\(\frac{L_1}{L_2} = \frac{5}{2}\)

According to the problem, this ratio is expressed as \(\frac{5}{x}\). Equating the two expressions gives:

\(\frac{5}{2} = \frac{5}{x}\)

By comparing, it is clear that \( x = 2 \).

Conclusion

Thus, the value of \( x \) is 2.