Question

The fundamental frequency of an air column in pipe 'A' closed at one end coincides with the second overtone of pipe 'B' open at both ends. The ratio of the length of pipe 'A' to that of pipe 'B' is

• A. 3 : 8
• B. 3 : 4
• C. 1 : 6
• D. 2 : 3

Hint:

Always be careful with terminology: for an open pipe, the $k^{\text{th}}$ overtone is always equal to the $(k+1)^{\text{th}}$ harmonic, so the second overtone is the third harmonic ($3 \times \text{fundamental}$). For a closed pipe, overtones skip directly across odd numbers, but here we only need the basic fundamental frequency ($1 \times \text{fundamental}$). Setting $\frac{1}{4L_A} = \frac{3}{2L_B}$ quickly yields the answer.

Answer 1

The Correct Option is C

Step 1: Write the fundamental of the closed pipe A.
A pipe closed at one end has a fundamental frequency $$n_A = \frac{v}{4L_A},$$ since only a quarter wavelength fits inside.
Step 2: Write the harmonics of the open pipe B.
A pipe open at both ends has fundamental $\dfrac{v}{2L_B}$ and produces all harmonics: $1f, 2f, 3f, \ldots$
Step 3: Identify the second overtone of B.
The first overtone is the 2nd harmonic and the second overtone is the 3rd harmonic, so $$n_B' = 3\times\frac{v}{2L_B} = \frac{3v}{2L_B}.$$
Step 4: Apply the coincidence condition.
The problem says $n_A = n_B'$, hence $$\frac{v}{4L_A} = \frac{3v}{2L_B}.$$
Step 5: Cancel $v$ and rearrange.
Dropping the common $v$, $$\frac{1}{4L_A} = \frac{3}{2L_B}\ \Rightarrow\ \frac{L_A}{L_B} = \frac{2}{4\times 3} = \frac{2}{12}.$$
Step 6: Simplify the ratio.
$$\frac{L_A}{L_B} = \frac{1}{6}.$$
\[ \boxed{L_A : L_B = 1 : 6} \]