
A pipe's lower end is immersed in water such that the length of the air column from the top open end has a certain length $25 \mathrm{~cm}$. The speed of sound in air is $350 \mathrm{~m} / \mathrm{s}$. The air column is found to resonate with a tuning fork of frequency $1750 \mathrm{~Hz}$. By what minimum distance should the pipe be raised in order to make the air column resonate again with the same tuning fork?
A. $7 \mathrm{~cm}$
B.$5~\text{cm}$
C.$35 \mathrm{~cm}$
D. $10 \mathrm{~cm}$
Answer
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Hint: An air column can be defined in a certain space as the weight or pressure of the air. Calculate The Initial length of the pipe and the minimum increase in length of the air column after increasing the length of the pipe by $\mathrm{Lm}$ after this Equate the equation by the frequency formula to get the minimum distance by the pipe.
Formula used:
$v_{\mathrm{n}}=\dfrac{(2 \mathrm{n}-1) \mathrm{v}}{4 \mathrm{~L}} \quad$
Complete answer:
Resonance describes the increased amplitude phenomenon that occurs when the frequency of a force applied periodically is equal to or close to the natural frequency of the system on which it acts.
Given: speed of sound in air, $\mathrm{v}=350 \mathrm{~m} / \mathrm{s}$
For one sided closed pipe:
Frequency, $v_{\mathrm{n}}=\dfrac{(2 \mathrm{n}-1) \mathrm{v}}{4 \mathrm{~L}} \quad$ where $\mathrm{n}=1,2,3$
When length of the pipe is $25 \mathrm{~cm}$ OR $0.25 \mathrm{~m}, \quad 1750=\dfrac{(2 \mathrm{n}-1) 350}{4 \times 0.25}$
$\text{n}=3$
Now let the length of the pipe after raising be $\mathrm{Lm}$.
For minimum increase in length of air column, $\mathrm{n}=3+1=4$
For minimum increase in length of air column, $\mathrm{n}=3+1=4$
$1750=\dfrac{(2\times 4-1)350}{4\times \text{L}}$
gives $\mathrm{L}=0.35 \mathrm{~m}$
Thus, the distance by which the pipe is raised $=0.35-0.25=0.1~\text{m}$
The distance by which the pipe is raised $=0.35-0.25=0.1 \mathrm{~m}=10 \mathrm{~cm}$
So,the Correct option is (D) .
Note:
Resonance describes the increased amplitude phenomenon that occurs when a periodically applied force is frequently applied. In a resonance column apparatus, air column vibrations can be set up. The condition for resonance is this. This occurs only when the air column length is proportional to one-fourth of the sound wavelength of sound waves that have a frequency equal to the tuning fork frequency.
Formula used:
$v_{\mathrm{n}}=\dfrac{(2 \mathrm{n}-1) \mathrm{v}}{4 \mathrm{~L}} \quad$
Complete answer:
Resonance describes the increased amplitude phenomenon that occurs when the frequency of a force applied periodically is equal to or close to the natural frequency of the system on which it acts.
Given: speed of sound in air, $\mathrm{v}=350 \mathrm{~m} / \mathrm{s}$
For one sided closed pipe:
Frequency, $v_{\mathrm{n}}=\dfrac{(2 \mathrm{n}-1) \mathrm{v}}{4 \mathrm{~L}} \quad$ where $\mathrm{n}=1,2,3$
When length of the pipe is $25 \mathrm{~cm}$ OR $0.25 \mathrm{~m}, \quad 1750=\dfrac{(2 \mathrm{n}-1) 350}{4 \times 0.25}$
$\text{n}=3$
Now let the length of the pipe after raising be $\mathrm{Lm}$.
For minimum increase in length of air column, $\mathrm{n}=3+1=4$
For minimum increase in length of air column, $\mathrm{n}=3+1=4$
$1750=\dfrac{(2\times 4-1)350}{4\times \text{L}}$
gives $\mathrm{L}=0.35 \mathrm{~m}$
Thus, the distance by which the pipe is raised $=0.35-0.25=0.1~\text{m}$
The distance by which the pipe is raised $=0.35-0.25=0.1 \mathrm{~m}=10 \mathrm{~cm}$
So,the Correct option is (D) .
Note:
Resonance describes the increased amplitude phenomenon that occurs when a periodically applied force is frequently applied. In a resonance column apparatus, air column vibrations can be set up. The condition for resonance is this. This occurs only when the air column length is proportional to one-fourth of the sound wavelength of sound waves that have a frequency equal to the tuning fork frequency.
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