Question

An open air pipe of length 80 cm has the second harmonic frequency equal to the fundamental frequency of a closed organ pipe. Find the length of the closed pipe.

• A. 20 cm
• B. 40 cm
• C. 60 cm
• D. 10 cm

Hint:

For open pipe, \(f_n = n v/2L\); for closed pipe, \(f_n = (2n-1)v/4L\). Equate harmonics as needed.

Answer 1

The Correct Option is A

Step 1: Write the two frequency conditions.
An open pipe of length $L_o$ has harmonics $f_n = \dfrac{nv}{2L_o}$, so its second harmonic is just $f_2 = \dfrac{v}{L_o}$. A closed pipe of length $L_c$ has only its fundamental, $f_1 = \dfrac{v}{4L_c}$.
Step 2: Match the two given frequencies.
We are told the open pipe's second harmonic equals the closed pipe's fundamental, so
\[ \frac{v}{L_o} = \frac{v}{4L_c} \]
The speed of sound cancels straight away, leaving a simple length relation: $4L_c = L_o$.
Step 3: Plug in the open pipe length.
With $L_o = 80\ \text{cm}$,
\[ L_c = \frac{80}{4} = 20\ \text{cm} \]
\[ \boxed{20\ \text{cm}} \]