WEBVTT ﻿1 00:00:02.089 --> 00:00:07.548 The Kundt’s tube experiment enables us to study stationary sound waves. 2 00:00:07.549 --> 00:00:09.455 Remember that a stationary wave 3 00:00:09.456 --> 00:00:14.261 is the sum of two progressive waves of equal frequency and amplitude 4 00:00:14.262 --> 00:00:17.109 but moving in opposite directions. 5 00:00:17.110 --> 00:00:19.538 The resultant resembles a lone vibration 6 00:00:19.539 --> 00:00:25.416 more than a wave but it is really a superposition of waves. 7 00:00:25.417 --> 00:00:27.714 Sound is a longitudinal wave 8 00:00:27.715 --> 00:00:31.006 that is, the displacements experienced by masses of air 9 00:00:31.007 --> 00:00:35.264 are parallel to the direction of the wave’s propagation. 10 00:00:35.265 --> 00:00:39.261 The curve “s of x, t” measures this displacement. 11 00:00:39.262 --> 00:00:40.985 Here we can read, in order 12 00:00:40.986 --> 00:00:45.608 the horizontal displacement around the equilibrium position of a layer of air 13 00:00:45.609 --> 00:00:47.829 along the x axis. 14 00:00:47.830 --> 00:00:50.728 For any ordinary frequency of excitation 15 00:00:50.729 --> 00:00:55.796 the curve “s of x, t” shows a disorderly motion of low amplitude. 16 00:00:55.797 --> 00:01:00.864 But, for certain frequencies, a stationary wave system sets in. 17 00:01:00.865 --> 00:01:05.122 The curve clearly shows the apppearance of antinodes of vibration 18 00:01:05.123 --> 00:01:09.667 areas where the air molecules vibrate with maximum amplitude 19 00:01:09.668 --> 00:01:14.604 and nodes of vibration, areas where they do not vibrate at all 20 00:01:14.605 --> 00:01:17.661 Note that the distance between two consecutive nodes 21 00:01:17.662 --> 00:01:21.892 is half the length of the wave: l/2 22 00:01:21.893 --> 00:01:25.550 Knowing the length of the tube and the resonant frequency, f 23 00:01:25.551 --> 00:01:31.296 we can determine the speed of sound, which is equal to l*f. 24 00:01:31.297 --> 00:01:37.331 For what is currently under observation, we measure half a wavelength as 25 cm 25 00:01:37.332 --> 00:01:40.596 and thus a wavelength is 50 cm. 26 00:01:40.597 --> 00:01:43.835 Since the frequency is 680 Hz, 27 00:01:43.836 --> 00:01:46.836 we find that the speed of sound is 340 meters per second.