Name: Standing Waves
Uploaded: 2019-07-19
Duration: 8 min 34 s
Description: Have you ever played the guitar? How does putting fingers on the string change the frequency of the sound produced? Curious? Watch the video and unravel the...

Question

A glass tube is closed at one end. The air column it contains has a length that can be varied between $0.50 m$ and $1.50 m$ . A tuning fork of frequency $306 Hz$ is sounded at the top of the tube. Predict the lengths of the air column at which loud sounds would be heard from the tube. (Take the speed of the sound to be $340 m s^{- 1}$ )

Accepted Answer

The loud sounds would be heard at the positions where the conditions for resonance are satisfied. The following cases are discussed to find the length of air column required for the loud sound:

(i) For the first resonance the length $(L_{0})$ of the air column in the resonance tube:

$λ = 4 L_{0}$

$\Rightarrow n_{0} = \frac{340}{4 L_{0}}$ ( using the equation $V = n λ$ )

$\Rightarrow L_{0} = \frac{340}{4 \times 306} = 0.27 m$

The required length of the tube does not lie between $0.50 m$ and $1.50 m$ hence, loud sound will not be heard.

(ii) For the first overtone or the second resonance which is the third harmonic:

$L_{1} = \frac{3 \times 340}{4 \times 306} = 0.81 m$

The required length of the air column lie between $0.50 m$ and $1.50 m$ hence, loud sound will be heard.

(iii) For the second overtone or the third resonce which is the fifth harmonic:

$L_{2} = 5 \times O . 27 = 1.35 m$

The required length of the air column lies between $0.50 m$ and $1.50 m$ hence, the loud sound would be heard.

(iv) For the third overtone or the fourth resonance which is the seventh harmonic:

$L_{3} = 7 \times 0.27 = 1.89 m$

The required length of the air column does not lie between $0.50 m$ and $1.50 m$ hence, loud sound will not be heard.

Hence, for the length of air columns equal to $0.81 m$ and $1.89 m$ the loud sound would be heard. It would not be heard for further higher overtones.

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