Question

An open pipe is in second harmonic with frequency \( f_1 \). One end of the tube is closed and frequency is increased to \( f_2 \) and it resonates again with nth harmonic. For what value of \( n \), the ratio of \( \frac{f_1}{f_2} \) is \( \frac{4}{5} \)?

• A. 3
• B. 9
• C. 5
• D. 7

Hint:

Open pipe supports all harmonics, closed pipe supports only odd harmonics.

Answer 1

The Correct Option is C

Step 1: Frequency in open pipe.
\[ f_n = \frac{nv}{2L} \]

Step 2: For second harmonic.
\[ f_1 = \frac{2v}{2L} = \frac{v}{L} \]

Step 3: Frequency in closed pipe.
\[ f_n = \frac{nv}{4L}, \quad n = 1,3,5,\dots \]

Step 4: Write second frequency.
\[ f_2 = \frac{nv}{4L} \]

Step 5: Form ratio.
\[ \frac{f_1}{f_2} = \frac{\frac{v}{L}}{\frac{nv}{4L}} \]

Step 6: Simplify.
\[ \frac{f_1}{f_2} = \frac{4}{n} \]

Step 7: Solve using given condition.
\[ \frac{4}{n} = \frac{4}{5} \Rightarrow n = 5 \]