Question

A pipe open at both ends produces a fundamental frequency \(n_1\). When the pipe is kept with \(\frac{3^{\text {th }}}{4}\) of its length in water, it produces a note of fundamental frequency \(\mathrm{n}_2\). The ratio of \(\frac{\mathrm{n}_1}{\mathrm{n}_2}\) is

• A. \(\frac{4}{3}\)
• B. \(\frac{3}{4}\)
• C. 2
• D. \(\frac{1}{2}\)

Hint:

Dipping three quarters of the open pipe in water leaves a closed pipe of one quarter the length.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
An open pipe of length \(L\) has fundamental frequency \(n_1=\dfrac{v}{2L}\).

Step 2: Pipe in water
With \(\dfrac34L\) in water, the air column is \(\dfrac L4\) and it is closed at one end.

Step 3: Closed pipe
\(n_2=\dfrac{v}{4(L/4)}=\dfrac vL\).

Step 4: Ratio
\(\dfrac{n_1}{n_2}=\dfrac{v/2L}{v/L}=\dfrac12\).

Final Answer:
The ratio \(n_1/n_2\) is 1/2. \[ \boxed{\text{(D) }\dfrac{1}{2}} \]