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
Show Hint
- $25 \%$
- $20 \%$
- $-20 \%$
- $-25 \%$
The Correct Option is A
Approach Solution - 1
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) $25 %$.
Approach Solution -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%.
\[ \boxed{\text{The fundamental frequency increases by } 25\%} \]
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