00:01
Hi there, so for this problem we have certain transfer sinusoidal wave of wavelength that is equal to 20 centimeters.
00:13
And it is moving in the positive direction of the x -axis.
00:18
So the transfer velocity of the particle at x equals to zero as a function of time is shown in here.
00:31
So where the scale of the vertical axis is said to be u .s is equal to 5 centimeters per second.
00:45
So for part a of this problem, we are asked, what is the wave speed? so to calculate that, we know that the frequency is equal to 1 over the period.
01:06
In the period, as you can see from the figure, is simply four seconds.
01:16
So if we substitute that in here, we have one over four seconds.
01:21
That will give us one over four hertz.
01:27
And the speed is simply the product between the frequency and the wavelength.
01:34
And we know those values because we know that the frequency is 1 over 4 hearths and the wavelength is equal to 20 centimeters.
01:50
So from this we obtain a speed of 5 centimeters per second.
01:58
Now for part b we are asked about the amplitude and we refer to the graph to see that the maximum transverse speed, which we will refer to as um, is c.
02:17
Five centimeters five centimeters per second that is the amplitude for the function that we are given now the simply harmonic motion relation is that the maximum value for the speed is equal to the maximum value for the position times the angular frequency we know that the angular frequency is defined as 2 times pi times the frequency.
02:54
So we just simply substitute those values in here.
02:56
We will have that this is 5 is equal to the amplitude times 2 pi divided by 1 over 4.
03:08
And if we solve for the amplitude of the position, this will give us 3 .2 centimeters.
03:15
So that's a solution for part b of this problem.
03:19
Now for par c, we are asked about the frequency...