(i) In a resonance tube experiment to determine the speed of sound in air, a pipe of diameter 5 cm is used.…
(i) In a resonance tube experiment to determine the speed of sound in air, a pipe of diameter 5 cm is used. The air column in pipe resonates with a tuning fork of frequency 480 Hz, when minimum length of air column is 16 cm. Find the speed of sound in air at room temperature.
(ii) A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. The velocity of sound in air is 320 m/s. Estimate the diameter of the tube. One end of the tube is now closed. Calculate the lowest frequency of resonance for the tube.
Solution
Sol. (i) Given, diameter, D = 5 cm
$\lambda/4 = l + e = l + 0.6r = l + 0.3D$ ($\therefore r = D/2$)
$\lambda = 4(l + 0.3D) = 4(16 + 0.3 \times 5)$
$= 70 cm = 0.7m$
Now, speed of sound, $v = f\lambda$
$= 480 \times 0.7 = 336 ms^{-1}$
(ii) Given, $l = 48$ cm, $f = 320$ Hz
Speed, $v = f\lambda \Rightarrow \lambda = v/f = 320/320 = 1m = 100 cm$
$\lambda/2 = l + 2e \Rightarrow \lambda = 2(l + 2e)$
$100 = 2(48 + 2e) \Rightarrow e = 1 cm$
$\Rightarrow 0.6r = 0.3D = 1$
$\Rightarrow D = 10/3 cm = 3.33 cm$
For closed pipe, $\lambda/4 = l + e = 48 + 1 = 49$
$\lambda = 196 cm = 1.96m$
$\therefore$ Speed, $v = f\lambda$
$\Rightarrow f = v/\lambda = 320/1.96 = 163.3 Hz$
Answer: 163.3 Hz