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Resonance Column Method - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
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26 Questions around this concept.
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A tuning fork of frequency 340 Hz resonates in the mth mode with an air column of length 125 cm in a cylindrical tube closed at one end. When water is slowly poured into it, the minimum height of water required for observing resonance once again is ________cm.
(Velocity of sound in air is $\left.340 \mathrm{~ms}^{-1}\right)$
In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning fork of frequency 400 Hz. The velocity of the sound at room temperature is $336 \mathrm{~ms}^{-1}$. The third resonance is observed when the air column has a length of $\qquad$ cm
In a resonance pipe, the first and second resonances are obtained at depths 22.7 cm and 70.2 cm respectively. what will be the end correction?
In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental modes, with a tuning fork is 0.1 m. when this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. calculate the end correction-
A tube closed at one end containing air produces, when excited, the fundamental note of frequency . If the tube is opened at both ends, the fundamental frequency
, that can be excited is -
Stationary waves of frequency are formed in a medium in which the velocity of sound is
. The distance between a node and the neighbouring antinode is :
A cylindrical resonance tube, open at both ends, has a fundamental in air. If half of the length is dipped vertically in water. The fundamental of the air column will be-
When a vibrating tuning fork of frequency 512 Hz is heard above the mouth of a resonance tube of adjustable length, the first two successive positions of resonance occur, when the length of the air columns are 15.4 cm and 48.6 cm respectively. Then, the velocity of sound is -
A glass tube of 1.0m length is filled with water can be drained out slowly at the bottom of the tube. If a vibrating tuning fork of frequency 500 c/s is brought at the upper end of the tube and the velocity of sound is 330 m/s, then the total number of resonance obtained will be-
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Use NowA tuning fork of $500 \mathrm{H}_2$ is used to produce resonance in a resonance tube experiment. The level of water at the first resonance is 35 cm and at second resonance is 70 cm.
The error in calculating the velocity of sound is(Take velocity of sound in air as $330 \mathrm{~m} / \mathrm{s}$ )
Concepts Covered - 1
Resonance column method
Resonance column method
In this the equipment used is resonance tube. This apparatus is used to determine the velocity of sound in air and used to compare frequency of two turning fork.
It is closed organ pipe with variable length of air column. When we brought a turning fork near it, its air column vibrates with the frequency of fork. The length of air column varied until the frequency of fork and the air column become equal. When frequency becomes equal, column resonates and the note become loud.
It is the full setup of resonance tube. If a tuning fork of known frequency $n_o$ is struck on a rubber pad and brought near to the open end. Because of this the air column starts oscillating. This air column behaves as a closed organ pipe and the water level as closed end. We decrease the water level gradually and as the water level reaches a position where there is a node of corresponding stationary wave, in air column, resonance takes place. At this place intensity of sound will be maximum.
Let at this position the length of air column is $l_1$. By further decreasing water level again after some distance maximum intensity of sound is obtained where the node is obtained. Let this level is $l_2$.
If $l_1$ and $l_2$ are the length of first and second resonance, then -
$$
\begin{aligned}
& l_1+e=\frac{\lambda}{4} \text { and } l_2+e=\frac{3 \lambda}{4} \\
& \text { so, } \lambda=2\left(l_2-l_1\right)
\end{aligned}
$$
Speed of sound in air at room temperature $v=n \lambda=2 n\left(l_2-l_1\right)$
Also,
$$
\begin{gathered}
\frac{l_2+e}{l_1+e}=3 \\
\Rightarrow l_2=3 l_1+2 e
\end{gathered}
$$
So, the second resonance is obtained at length more than thrice the length of first resonance.
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