A tuning fork of frequency 340 , Hz is vibrat | vedclass

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(C) The wavelength of the sound wave is given by $\lambda = \frac{v}{
u}$.Substituting the values,$\lambda = \frac{340}{340} = 1 \, m = 100 \, cm$.Fo

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) The wavelength of the sound wave is given by $\lambda = \frac{v}{
u}$.Substituting the values,$\lambda = \frac{340}{340} = 1 \, m = 100 \, cm$.For a closed organ pipe (resonance tube),resonance occurs when the length of the air column $l$ satisfies $l = \frac{(2n-1)\lambda}{4}$ for $n = 1, 2, 3, \dots$.Possible lengths of the air column are:For $n=1$: $l_1 = \frac{\lambda}{4} = \frac{100}{4} = 25 \, cm$.For $n=2$: $l_2 = \frac{3\lambda}{4} = \frac{300}{4} = 75 \, cm$.For $n=3$: $l_3 = \frac{5\lambda}{4} = \frac{500}{4} = 125 \, cm$.Since the total length of the tube is $120 \, cm$,the air column length $l_3 = 125 \, cm$ is not possible.As water is poured,the air column length decreases. The first resonance occurs at $l_2 = 75 \, cm$ (as $l_1 = 25 \, cm$ would require more water,and we seek the minimum height of water for the first possible resonance encountered).The height of the water column $h$ is given by $h = 	ext{Total length} - 	ext{Air column length} = 120 \, cm - 75 \, cm = 45 \, cm$.

$A$ tuning fork of frequency $340 \, Hz$ is vibrated just above a cylindrical tube of length $120 \, cm$. Water is slowly poured into the tube. If the speed of sound is $340 \, m/s$,then the minimum height of water required for resonance is .... $cm$.

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