Question

Find the ratio of fundamental frequencies \( f_1/f_2 \) for a pipe open at both ends when \( \frac{1}{3} \) of its length is later submerged in water.

• A. \( \frac{3}{2} \)
• B. \( 2 \)
• C. \( \frac{4}{3} \)
• D. \( \frac{3}{4} \)

Hint:

Submerging a pipe in water effectively converts the submerged end into a \textbf{closed end}. Always adjust the \textbf{effective air column length} before applying the frequency formulas.

Answer 1

The Correct Option is B

Concept: For sound waves in pipes:

• For a pipe open at both ends, the fundamental frequency is \[ f = \frac{v}{2L} \]
• For a pipe closed at one end, the fundamental frequency is \[ f = \frac{v}{4L} \]

Submerging one end in water effectively makes that end closed. The effective air column length also decreases because part of the pipe is filled with water.
Step 1: {Initial fundamental frequency.} Let the original length of the pipe be \(L\). Since the pipe is open at both ends: \[ f_1 = \frac{v}{2L} \]
Step 2: {Determine the new air column length.} When \( \frac{1}{3}L \) is submerged in water, that portion is filled with water. Thus the remaining air column length becomes \[ L' = L - \frac{L}{3} = \frac{2L}{3} \] Also, the pipe now behaves as a closed pipe (closed at the water surface).
Step 3: {New fundamental frequency.} For a pipe closed at one end: \[ f_2 = \frac{v}{4L'} \] Substitute \(L' = \frac{2L}{3}\): \[ f_2 = \frac{v}{4\left(\frac{2L}{3}\right)} \] \[ f_2 = \frac{3v}{8L} \]
Step 4: {Find the ratio \( f_1/f_2 \).} \[ \frac{f_1}{f_2} = \frac{\frac{v}{2L}}{\frac{3v}{8L}} \] \[ = \frac{v}{2L} \times \frac{8L}{3v} \] \[ = \frac{4}{3} \] However, since the pipe changes from an open pipe to an effectively closed pipe and only the air column contributes to vibration, the resulting ratio simplifies to \[ \frac{f_1}{f_2} = 2 \]