(B) The tube acts as a closed organ pipe because one end is closed by the water surface and the other is open. The resonance condition for a closed pi

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(B) The tube acts as a closed organ pipe because one end is closed by the water surface and the other is open. The resonance condition for a closed pipe is given by $l = \frac{(2N - 1)v}{4f}$,where $N$ is the order of the resonance,$v = 330 \ m/s$ is the speed of sound,and $f = 660 \ Hz$ is the frequency.Substituting the values: $1.5 = \frac{(2N - 1) \times 330}{4 \times 660}$.Simplifying the equation: $1.5 = \frac{(2N - 1)}{8}$.$12 = 2N - 1$.$2N = 13$.$N = 6.5$.Since $N$ must be an integer,the possible values for $N$ are $1, 2, 3, 4, 5, 6$. Thus,there are $6$ resonances.

Class 11 Physics Waves and Sound Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

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$A$ glass tube $1.5 \ m$ long and open at both ends is immersed vertically in a water tank completely. $A$ tuning fork of $660 \ Hz$ is vibrated and kept at the upper end of the tube,and the tube is gradually raised out of the water. Taking the velocity of sound in air as $330 \ m/s$,find the total number of resonances heard before the tube comes out of the water.

A
$12$
B
$6$
C
$8$
D
$4$

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Class 11

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Waves and Sound

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Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

$A$ source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude ${\rho _0}$ inside the tube. If the atmospheric pressure is ${\rho _A},$ then the ratio of maximum and minimum pressure at the closed end of the tube will be

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In an open-end organ pipe of length $L$,if the velocity of sound is $V$,then the fundamental frequency will be (Neglect end correction).

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An open and a closed organ pipe are of same length. The ratio of the frequency of the $n^{th}$ mode of vibration of the two pipes is

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$A$ pipe's lower end is immersed in water such that the length of the air column from the top open end has a length of $25 \, cm$. The speed of sound in air is $350 \, m/s$. The air column is found to resonate with a tuning fork of frequency $1750 \, Hz$. By what minimum distance should the pipe be raised in order to make the air column resonate again with the same tuning fork?

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$A$ source of frequency $340 \,Hz$ is kept above a vertical cylindrical tube closed at the lower end. The length of the tube is $120 \,cm$. Water is slowly poured in just enough to produce resonance. Then, the minimum height (velocity of sound $= 340 \,m/s$) of the water level in the tube for that resonance is (in $\,m$)

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$A$ source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude ${\rho _0}$ inside the tube. If the atmospheric pressure is ${\rho _A},$ then the ratio of maximum and minimum pressure at the closed end of the tube will be

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