Question

In an experiment with a closed organ pipe, it is filled with water by $\left(\frac{1}{5}\right)$ th of its volume. The frequency of the fundamental note will change by

• A. $25 \%$
• B. $20 \%$
• C. $-20 \%$
• D. $-25 \%$

Hint:

The frequency of the fundamental note changes with the length of the pipe.

Answer 1

The Correct Option is A

1. Initial frequency: \[ f_1 = \frac{v}{4l} \]
2. Frequency with water: \[ f_2 = \frac{v}{4(4l/5)} = \frac{5v}{16l} \]
3. Percentage change: \[ \frac{\Delta f}{f} = \frac{\frac{5v}{16l} - \frac{v}{4l}}{\frac{v}{4l}} \times 100 = 25% \] Therefore, the correct answer is (1) $25 %$.

Answer 2

We are given that a closed organ pipe is partially filled with water by \( \frac{1}{5} \) of its volume. We are to find how the frequency of its fundamental note changes.

Concept Used:

The fundamental frequency of a closed organ pipe (one end closed, one end open) is given by:

\[ f = \frac{v}{4L} \]

where \( v \) is the speed of sound in air and \( L \) is the effective length of the air column. When water fills part of the pipe, the effective length of the air column decreases, and hence the frequency increases because \( f \propto \frac{1}{L} \).

Step-by-Step Solution:

Step 1: Let the total length of the pipe be \( L \).

Since the pipe is filled with water by \( \frac{1}{5} \) of its volume, the air column length becomes:

\[ L' = L - \frac{L}{5} = \frac{4L}{5} \]

Step 2: Write the new frequency of the fundamental note after filling with water.

\[ f' = \frac{v}{4L'} \] \[ f' = \frac{v}{4 \times \frac{4L}{5}} = \frac{5v}{16L} \]

Step 3: Write the ratio of the new frequency to the original frequency.

\[ \frac{f'}{f} = \frac{\frac{5v}{16L}}{\frac{v}{4L}} = \frac{5}{16} \times 4 = \frac{5}{4} \]

Step 4: Hence, the new frequency is \( \frac{5}{4} \) times the original frequency.

Final Computation & Result:

The frequency increases by a factor of \( \frac{5}{4} \), i.e. by 25%.

\[ \boxed{\text{The fundamental frequency increases by } 25\%} \]