Question

A pipe opened at both end produces a note of frequency ${f_1}$ .when the pipe is keep with $\dfrac{3}{4}th$ of its length in water it produces a note of frequency ${f_2}$ .the ratio $\dfrac{{{f_1}}}{{{f_2}}}$ is.
a. $\dfrac{3}{4}$
b. $\dfrac{4}{3}$
c. $\dfrac{1}{2}$
d. $2$

Answer 1

To solve this question we compare the length of pipe and wavelength of produced fundamental wave for both open pipe and closed pipe. By this we will find wavelength of wave in terms of length of pipe.
And after that we use the relation between speed of wave in pipe and frequency and wavelength of produced wave $v = f \times \lambda$.

Complete step by step answer:
Step 1:

We know there is always an antinode at the open end as shown in figure.
We take the length of the open organ pipe is L and the wavelength of produced wave is ${\lambda _1}$ and frequency produced is ${f_1}$
Then from the figure we can clearly see that there are antinodes (A) at the two ends and a node (N) in the middle of the pipe.
Then, From figure we can see $L = \dfrac{{{\lambda _1}}}{4} + \dfrac{{{\lambda _1}}}{4}$
$\Rightarrow L = \dfrac{{{\lambda _1}}}{2}$
$\Rightarrow {\lambda _1} = 2L$
We know $v = f \times \lambda$
Where $v \Rightarrow$ velocity of wave
$f \Rightarrow$ frequency of note
$\lambda \Rightarrow$ wavelength of note
Applying this,
$\Rightarrow v = {f_1} \times {\lambda _1}$
$\Rightarrow {f_1} = \dfrac{v}{{{\lambda _1}}}$
Put the value of ${\lambda _1}$
$\Rightarrow {f_1} = \dfrac{v}{{2L}}$ .................... (1)

Step 2:

when the pipe is keep with $\dfrac{3}{4}th$ of its length into water then the one end of pipe will close then at the open end there formed a anitinode (A) and at the closed end node (N) formed as shown in figure
from figure we can clearly see that $\dfrac{L}{4}$ length of the pipe is outside the water
so $\dfrac{L}{4} = \dfrac{{{\lambda _2}}}{4}$
${\lambda _2} = L$
Again apply formula $v = f \times \lambda$
$\Rightarrow v = {f_2} \times {\lambda _2}$
$\Rightarrow {f_2} = \dfrac{v}{{{\lambda _2}}} \\\ \Rightarrow {f_2} = \dfrac{v}{L} \\\$ put the value of ${\lambda _2}$
So we get
$\Rightarrow {f_2} = \dfrac{v}{L}$ ................ (2)

Step 3:
Now divided equation (1) by (2)
$\Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{v}{{2L}} \times \dfrac{L}{v}$
Further solving
$\therefore \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{1}{2}$
Hence we get $\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{1}{2}$

Hence, the correct answer is option (C).

Note: This is another method to solve this question.
We can solve it by a shorter method if you remember the fundamental tone of open pipe is given by
${f_1} = \dfrac{v}{{2L}}$ ....... (1)
And the fundamental tone for closed pipe
${f_2} = \dfrac{v}{{4L}}$, Here $L = \dfrac{L}{4}$
${f_2} = \dfrac{v}{L}$ ......... (2)
Divide (1) by (2)
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{v}{{2L}} \times \dfrac{L}{v} \\\ \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{1}{2} \\\$