Question 343: Waves

Question

The fundamental frequency of an air column in a pipe open at both ends is ' $\mathrm{f}_1$ '. Now $80 \%$ of its length is immersed in water, the fundamental frequency of the air column becomes $f_2$. The ratio of $f_1: f_2$ is

Accepted Answer

The fundamental frequency of an air column in a pipe open at both ends is determined by the equation:
$$ f_1 = \frac{v}{2L} $$
where $ v $ is the speed of sound in air and $ L $ is the length of the air column.
When 80% of the pipe is immersed in water, only 20% of the length remains filled with air and acts as an open pipe, since the other end of the air column is now effectively 'closed' by the water surface. Therefore, the effective length $ L' $ of the air column is:
$$ L' = 0.2L $$
For an open-closed column (since the top is open and the bottom is closed by water), the fundamental frequency is given by:
$$ f_2 = \frac{v}{4L'} $$
Substituting $ L' = 0.2L $ into the equation for $ f_2 $:
$$ f_2 = \frac{v}{4(0.2L)} = \frac{v}{0.8L} = \frac{5v}{4L} $$
Now, to find the ratio $ f_1: f_2 $:
$ \frac{f_1}{f_2} = \frac{\frac{v}{2L}}{\frac{5v}{4L}} = \frac{v}{2L} 	imes \frac{4L}{5v} = \frac{4}{10} = \frac{2}{5} $
The ratio $ f_1: f_2 $ is therefore $ 2:5 $.
Option D: $ 2:5 $ is the correct answer.

Answer

$5: 2$

Answer

$4: 5$

Answer

$5: 4$

Answer

$2: 5$

Waves

The fundamental frequency of an air column in a pipe open at both ends is ' \mathrm{f}_1 '. Now 80 \% of its length is immersed in water, the fundamental frequency of the air column becomes f_2. The ratio of f_1: f_2 is

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