Question

A simple apparatus for demonstrating resonance in an air column is depicted in figure. A vertical pipe open at both ends is partially submerge in water, and a tuning fork vibration at an unknown frequency is placed near the top of the pipe. The length $L$ of the air column can be adjusted by moving the pipe vertically. The sound waves generated by the fork are reinforced of the pipe. For a certain pipe, the smallest value of $L$ for which a peak occurs in the sound intensity is $9.00 cm .$

Question Image

$(a)$ What is the frequency of the tuning fork?
$(b)$ What are the values of $L$ for the next two resonance conditions?

Accepted Answer

Resonance column tube behaves like close organ pipe and fundamental frequency is given by,

The fundamental frequency for $L = 0.090 m$

$f_{1} = \frac{v}{4 L} = \frac{343 m / s}{4 (0.090 m)} = 953 Hz$

Now the wavelength of wave by tuning fork can be determined by using relation $v = λ f$

$λ = \frac{v}{f} = \frac{343 m / s}{953 Hz} = 0.360 m$

The length of the air column for the second resonance is $3 λ / 4$

$L = 3 λ / 4 = 0.270 m$

Next length of the air column for the third resonance is $5 λ / 4 .$
$L = 5 λ / 4 = 0.450 m$

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