Question

The frequency of fifth harmonic of a closed pipe is equal to the frequency of third harmonic of an open pipe. If the length of the open pipe is 72 cm, then the length of the closed pipe is:

• A. 60 cm
• B. 45 cm
• C. 30 cm
• D. 75 cm

Hint:

When solving for the lengths of pipes with different harmonic frequencies, use the relationship between the fundamental frequencies of open and closed pipes to set up a ratio.

Answer 1

The Correct Option is A

To determine the length of the closed pipe, we need to compare the frequencies of the harmonics of both the closed and open pipes. Given:
- Frequency of the fifth harmonic of a closed pipe equals the frequency of the third harmonic of an open pipe.
- Length of the open pipe, \( L_{\text{open}} = 72 \, \text{cm} \). Step 1: Determine the Frequencies For a closed pipe, the frequency of the \( n \)
-th harmonic is: \[ f_{\text{closed}, n} = \frac{n v}{4 L_{\text{closed}}} \] where \( n \) is an odd integer (1, 3, 5, ...). For an open pipe, the frequency of the \( n \)
-th harmonic is: \[ f_{\text{open}, n} = \frac{n v}{2 L_{\text{open}}} \] where \( n \) is any positive integer (1, 2, 3, ...). Step 2: Set the Frequencies Equal Given that the fifth harmonic of the closed pipe equals the third harmonic of the open pipe: \[ f_{\text{closed}, 5} = f_{\text{open}, 3} \] Substitute the expressions: \[ \frac{5 v}{4 L_{\text{closed}}} = \frac{3 v}{2 L_{\text{open}}} \] Step 3: Solve for \( L_{\text{closed}} \) Cancel \( v \) from both sides: \[ \frac{5}{4 L_{\text{closed}}} = \frac{3}{2 L_{\text{open}}} \] Rearrange to solve for \( L_{\text{closed}} \): \[ L_{\text{closed}} = \frac{5 \times 2 L_{\text{open}}}{4 \times 3} = \frac{10 L_{\text{open}}}{12} = \frac{5 L_{\text{open}}}{6} \] Substitute \( L_{\text{open}} = 72 \, \text{cm} \): \[ L_{\text{closed}} = \frac{5 \times 72}{6} = \frac{360}{6} = 60 \, \text{cm} \] Final Answer: \[ \boxed{60 \, \text{cm}} \] This corresponds to option (1).