Question

The length of an open pipe is half of the length of another closed pipe. When the two pipes are vibrated, four nodes are formed in both the cases. The ratio of the frequencies of the open and the closed pipes is:

• A. $1:1$
• B. $16:7$
• C. $16:3$
• D. $7:3$

Hint:

For an open organ pipe having $N$ nodes, \[ L=N\left(\frac{\lambda}{2}\right). \] For a closed organ pipe having $N$ nodes, \[ L=(2N-1)\left(\frac{\lambda}{4}\right). \] After finding the wavelengths, use \[ f=\frac{v}{\lambda} \] to obtain the frequency ratio quickly without calculating the actual frequencies.

Answer 1

The Correct Option is B

Step 1: Set up the pipe lengths.
Let the open pipe have length $L_o$ and the closed pipe length $L_c$. We are told $L_o = \tfrac{L_c}{2}$, that is $L_c = 2L_o$.
Step 2: Use the node count for the open pipe.
In an open pipe both ends are antinodes. For four nodes inside, the open pipe vibrates in its fourth harmonic, so its frequency is $f_o = \dfrac{4v}{2L_o} = \dfrac{2v}{L_o}$.
Step 3: Use the node count for the closed pipe.
A closed pipe supports only odd harmonics. The harmonic that produces four nodes corresponds to $n = 7$, so its frequency is $f_c = \dfrac{7v}{4L_c}$.
Step 4: Substitute the length relation.
Replace $L_c = 2L_o$: $f_c = \dfrac{7v}{4(2L_o)} = \dfrac{7v}{8L_o}$.
Step 5: Form the ratio.
$\dfrac{f_o}{f_c} = \dfrac{2v/L_o}{7v/8L_o} = \dfrac{2v}{L_o}\times \dfrac{8L_o}{7v} = \dfrac{16}{7}$.
Step 6: Conclude.
The frequency ratio of the open to the closed pipe is $16:7$, which is option (2).
\[ \boxed{f_o : f_c = 16 : 7} \]