velocity of sound

How to Measure the Velocity of Sound in a Quincke Tube

Quincke devised a simple method of measuring the velocity of sound by obtaining permanent interference between two sound waves. He used a closed tube SAEB which had openings at S, E, and placed a source of sound at S. A wave then traveled in the direction SAE round the tube, while another wave traveled in the opposite direction SBE; and since these waves are due to the same source, S, they always set out in phase, i.e., they are coherent.

Like a trombone, one side, b, of the tube can be pulled out, thus making SAE, SBE of different lengths. When SAE and SBE are equal in length an observer as E hears a loud sound, since the paths of the two waves are then equal. As B is pulled out the sound dies away and becomes a minimum when the path difference, SBE-SAE, is λ/2, where λ is the wavelength. In this case the two waves arrive 1800 out of phase. If the tube is pulled out farther, the sound increases in loudness to a maximum; the path difference is then λ. If k is the distance moved from one position of minimum sound, MN say, to the next position of minimum sound, PQ say, then 2k = λ. Thus the wavelength of the sound can be simply obtained by measuring k.

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The velocity of sound in the tube is given v = f λ, where f is the frequency of the source s, and thus v can be found when a source of known frequency is used. In a particular experiment with Qunicke’s tube, the tube B was moved a distance 4.28cm between successive minima of sound, and the frequency of the source was 400 Hz.

Thus λ = 2 x 4.28cm

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and v = fλ = 4000 x 2 x 4.28 = 34240cm s-1 = 342.4ms-1

It can be seen that, unlike reflection and refraction, the phenomenon of interference can be utilized to measure the wavelength of sound waves and the velocity of sound. We shall see later that interference is also utilized to measure the wavelength of high waves.

Destructive Interference

Consider now a point P whose distance from B is half a wavelength longer than its distance from A, i.e., AP-BP = λ/2. The vibration at P due to B will then be 1800 out of phase with the vibration there to A. the resultant effect at P is thus zero, as the displacements at any instant are equal and opposite to each other, no sound is therefore heard at P, and the permanent silence is said to be due to “destructive interference” between the sound waves from A and B.

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If the path difference, AP – BP, were 3λ/2 or 5 λ/2, instead of λ/2, permanent silence would also exist at P as the vibrations, there due to A, B would again be 1800 out of phase. Summarizing, then,

Silence occurs if the path-difference is an odd number of half wavelengths, and

A loud sound occurs if the path- difference is a whole number of wavelengths.

The total sound energy in all the positions of loud sound discussed above is equal to the total sound energy of the two sources A, B, from the principle of the conservation of energy. The extra sound at the positions of loud sound thus makes up for the absent sound in the positions of silence.

One comment

  1. why is the wavelength x2 as surely the distance between two minima of sound is equal to the wavelength as it is two consecutive points on the same part of the wave? also you quote the frequency to be 400Hz and then use 4000Hz in the calculation.
    Thanks in advance and thanks for a great explanation otherwise

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