A tuning fork of frequency 340 Hz is vibrate | vedclass

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(A) The tuning fork is in resonance with an air column in a pipe closed at one end. The resonant frequency is given by $n = \frac{(2N-1)v}{4l}$,where

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$45$

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(A) The tuning fork is in resonance with an air column in a pipe closed at one end. The resonant frequency is given by $n = \frac{(2N-1)v}{4l}$,where $N = 1, 2, 3, \dots$ corresponds to different modes of vibration.Substituting $n = 340 \ Hz$ and $v = 340 \ m/s$ (speed of sound in air),the length of the air column $l$ is:$l = \frac{(2N-1)v}{4n} = \frac{(2N-1) 	imes 340}{4 	imes 340} = \frac{2N-1}{4} \ m = (2N-1) 	imes 25 \ cm$.For $N = 1, 2, 3, \dots$,the possible lengths of the air column are $l = 25 \ cm, 75 \ cm, 125 \ cm, \dots$.Since the tube is only $120 \ cm$ long,the possible lengths of the air column are $25 \ cm$ and $75 \ cm$.The corresponding height of the water column is $h = 	ext{Total height} - l$.For $l = 25 \ cm$,$h = 120 - 25 = 95 \ cm$.For $l = 75 \ cm$,$h = 120 - 75 = 45 \ cm$.The minimum height of water required for resonance is $45 \ cm$.

$A$ tuning fork of frequency $340 \ Hz$ is vibrated just above a tube of $120 \ cm$ height. Water is slowly poured in the tube. What is the minimum height of water necessary for resonance (in $cm$)?

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