A string with both ends fixed is vibrating at its 3rd...

A string with both ends fixed is vibrating at its 3rd harmonic. The waves have a speed 192 m / s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.4 cm. a) Calculate the amplitude of motion of points on the string at a distance of i) 40.0 cm; ii) 20.0 cm; iii) 10.0 cm from the left end of the rope. b) In each of the points in part a), how long it takes the rope to go from its greatest upward displacement to its greatest displacement down? c) Calculate the maximum transverse velocity and acceleration of the rope at each of the points of part a).

Transcript

00:01 For this problem, we're considering the third harmonic.

00:05 So we have three antinodes and four nodes for a wave with a speed of 192 meters per second, a frequency of 240 hertz, and an amplitude of 0 .4 centimeters or 0 .004 meters.

00:21 And we're asked firstly to calculate the amplitude of the motion at various locations along the length of the structure.

00:30 So to do this, i want to use the fact that the sort of maximum displacement x at any one location is given by a sign of omega -t.

00:49 And that t is really just showing how the wave is traveling over time, but i'm going to use this to sort of understand how the amplitude is going to change over time at a certain location.

01:02 And really what we need to do is figure out what to plug in here for omega -t based off of what position we are on our string.

01:14 So firstly, we need to know what the full length of the string is, and we can do that by first finding the wavelength.

01:23 The wavelength can be found using the wave speed equation, which is that wave speed is equal to wavelength times frequency.

01:30 So the wavelength is going to be wave -spread.

01:33 Speed divided by frequency and ends up being a value of 0 .8 meters.

01:39 And looking at this wave, i can see that there are three halves wavelength over the full length of the string.

01:47 So the length is three halves times the wavelength or 1 .2 meters.

01:55 So if on the left hand side we are at zero meters, on the right hand side we're at 1 .2.

02:02 And we can split up the wave from there.

02:06 We're at 0 .6 in the middle here, which makes this node 0 .4 and this node 0 .8, this antinode will be 0 .2, and this anti node will be 1.

02:24 Now i want to relate that to sort of the angle, where we start with 0 at the left.

02:32 One full wavelength is, going to be pi or sorry not pie two pi half of a wavelength is pie a quarter of a wavelength is pie over two we'll have three halves pie here and so on and so forth so now i think we're at a good position to actually solve the problem for part one we want to find the amplitude at 40 centimeters.

03:05 So 40 centimeters is at a location of pi on our standing wave here.

03:11 And so x is going to be equal to the amplitude times the sign of pi, which pi, sign of pi radians is going to be equal to zero, which makes sense because that is a node.

03:28 We can repeat this for our second location, which is 20 centimeters.

03:36 20 centimeters corresponds to pi over 2.

03:41 So we have x is equal to a sign of pi over 2.

03:47 And sign of pi over 2 is 1, so we end up with a value of 0 .004 for 20 centimeters.

04:00 And then lastly, for 10 centimeters, that's going to be pi over four.

04:08 And so we'll plug in again to a sign of pi over four.

04:13 And this time we get something that is not quite as nice...