Question
Question: A tuning fork vibrates at a frequency of 800 Hz and produces resonance in a resonance column tube. T...
A tuning fork vibrates at a frequency of 800 Hz and produces resonance in a resonance column tube. The upper end is left open and the lower end is closed by a water surface which can be varied. Successive resonances are observed at lengths 9.75 cm, 31.25 cm and 52.75 cm. With the help of above data, obtain the speed of sound in air.
Solution
Firstly, we consider the successive resonance lengths as l1,l2,l3 and then these are substituted in the formula of resonant frequency, v=4lnu. Three equations (i),(ii),(iii) are obtained. Then by solving them, expressions for l3−l2,l2−l1, equation (iv) are deduced, further values of l3−l2,l2−l1 from experimental data given in question are obtained which are found to be equal. Now the obtained values are equated with the expression (iv). Thus, by simple calculation value of u,i.e., speed of sound in air is calculated.
Complete step by step solution:
Consider successive resonance lengths as l1,l2 and l3 such that, l1=9.75cm, l2=31.25cm and l3=52.75cm.
Now, for the resonance column tube open at one end,
the resonant frequency, v=4lnu
where,nis an odd-positive integer,
and,v is the resonance frequency
Suppose the frequency of tuning fork is ν and l1,l2 and l3 are the successive lengths of tube in resonance with it, then, we have,
4l1nu=ν …(i)
4l2(n+2)u=ν...(ii)
4l3(n+4)u=ν...(iii)
From (iii)−(ii)and (ii)−(i), we get,
l3−l2=l2−l1=2νu...(iv)
According to given data, l3−l2=(52.75−31.25)cm=l2−l1=(31.25−9.75)cm=21.50cm
Now, from eq. (iv),
2vu=21.50cm
Or, u=2v×21.50cm=2×800s−1×21.50cm=344ms−1
Thus, speed of air as calculated from the given data is 344ms−1.
Note: The above method used for determination of speed of sound is referred to as ‘Resonance Column Method’. Resonance Column apparatus is used to measure the speed of sound in the air. It is one of the simplest techniques to determine the speed of sound by taking the note of harmonics.