Question

The air columns in two tubes closed at one end vibrating in their fundamental modes produce 2 beats per second. The number of beats produced per second when the same tubes are vibrated in their fundamental mode with their both ends open are

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For a pipe of a given length $L$: - Fundamental frequency of a closed pipe: $f_c = v/4L$. (Only odd harmonics are present). - Fundamental frequency of an open pipe: $f_o = v/2L$. (All harmonics are present). A simple relationship to remember is $f_o = 2f_c$.

Answer 1

The Correct Option is D

Step 1: Write the formula for the fundamental frequency of a closed pipe.
A tube closed at one end (a closed organ pipe) supports standing waves where the closed end is a node and the open end is an antinode. The fundamental frequency ($f_c$) for a pipe of length $L$ is given by: \[ f_c = \frac{v}{4L}, \] where $v$ is the speed of sound in air.

Step 2: Use the given beat frequency information for the closed pipes.
Let the two tubes have lengths $L_1$ and $L_2$, and their fundamental frequencies be $f_{c1}$ and $f_{c2}$. \[ f_{c1} = \frac{v}{4L_1}, f_{c2} = \frac{v}{4L_2}. \] The number of beats per second is the difference in their frequencies: \[ |f_{c1} - f_{c2}| = 2 \text{ Hz}. \]

Step 3: Write the formula for the fundamental frequency of an open pipe.
A tube open at both ends (an open organ pipe) supports standing waves where both ends are antinodes. The fundamental frequency ($f_o$) for a pipe of length $L$ is given by: \[ f_o = \frac{v}{2L}. \]

Step 4: Relate the fundamental frequencies of open and closed pipes.
For a pipe of the same length $L$, we can see the relationship between the two types of fundamental frequencies: \[ f_o = \frac{v}{2L} = 2 \left(\frac{v}{4L}\right) = 2f_c. \] The fundamental frequency of an open pipe is twice the fundamental frequency of a closed pipe of the same length.

Step 5: Calculate the new beat frequency for the open pipes.
When both tubes are open at both ends, their new fundamental frequencies will be $f_{o1}$ and $f_{o2}$. \[ f_{o1} = 2f_{c1}, f_{o2} = 2f_{c2}. \] The new beat frequency will be the difference between these new frequencies: \[ \text{Beats}_{new} = |f_{o1} - f_{o2}| = |2f_{c1} - 2f_{c2}|. \] \[ \text{Beats}_{new} = 2|f_{c1} - f_{c2}|. \] Substituting the original beat frequency: \[ \text{Beats}_{new} = 2 \times (2 \text{ Hz}) = 4 \text{ Hz}. \] The number of beats produced per second is 4.