0:00
So this is a bit lengthy.
00:01
We should write our givens.
00:03
So our v equals rather a speed of propagation equals 192 meters per second.
00:09
Our frequency is going to be 240 hertz.
00:16
Our amplitude is going to be 4 centimeters or rather 0 .4 centimeters or 0 .004 meters.
00:27
And then we can find the wavelength because that is simply the velocity divided by the frequency.
00:38
It's going to be 192 over 240 or 0 .8 meters.
00:46
And then we can find the angular frequency by simply multiplying the frequency by 2 pi.
00:55
It's going to be 480 pi radians per second.
00:59
So that'll be your angular frequency, our wavelength, and all of our givens.
01:07
So it's trying to say, okay, what's asking us, rather, what's the amplitude at an x coordinate of 0 .4 meters, x coordinate of 0 .2 meters, an x coordinate of 0 .1 meters.
01:21
So let's find the general equation for position y on a transverse wave, or rather a standard, a standing wave equation.
01:31
Of a on a string, y equals x of t and this is going to be a times cosine of omega t times sine of kx, k being again the weight, k being the wave number rather.
01:51
And at this point, we are saying that this is going to be evaluated at t equals zero second.
01:57
So we'll just say that we're going to evaluate the amplitude at these positions.
02:03
However, we're evaluating the amplitude at the positions initially.
02:08
So it's initially, if it's initially, then that means that t there's no time elapsed, and t can be zero seconds.
02:17
So if t is going to be zero seconds, this is going to become one, and this will effectively eliminate this term.
02:25
So we can say that y is only dependent on x, and we can say that this is going to be equal to a sign of, kx.
02:36
At this point, we can say y at x equals 0 .4 meters, y at x equals 0 .2 meters, and then y at x equals 0 .1 meter.
02:50
And then you simply have to substitute all of this into the equation here.
02:57
So we'll have 0 .004 for all of them.
03:03
This is simply just to get them all in one one go essentially so sign sign sign and then the wave the wave number the wave number is simply going to be equal to k the two pi over lambda and if we were to solve for this it would be two pi over point eight so we're saying that k is going to be two point five is going to be two point five so we'll have 2 .5 pi and then this will be times our x coordinates .4 .2 .1.
03:56
And all you have to do is plug all this into a calculator and we know that this is going to be 0 meters, 0 .004 meters and then somewhere in between 0 .00283 meters.
04:14
So this will be the amplitude at each of these exposition's, x positions.
04:20
And k, we have found k the wave number by simply finding, rather simply dividing 2 pi by the wavelength lambda.
04:30
So the next question is asking, what is the time taken to go from the largest upward displacement to the largest downward displacement? and this at any particular point, largest upward displacement to largest downward displacement, this is a half cycle.
05:02
So for any particular point and half cycle, is this going to be a half period.
05:10
And at this point, we can just say 1 over 2, t, and then we'll say 1 over 2, 1 over f, because the period is simply reciprocal of the frequency.
05:29
And then we can have 1 over 2.
05:33
The frequency is 240.
05:35
So we'll have 1 over 480 hertz.
05:42
And this is going to be 0 .0028 seconds.
05:47
So that will be how long, that will be the time taken to go from the largest upward displacement to the largest downward displacement, regardless of your position.
05:57
It does not matter what your position is.
06:00
It will always be 0 .00 -208 seconds for this value.
06:06
Now it's asking us to find the maximum transverse velocity at those three given points.
06:14
So let's bring up a new worksheet.
06:16
And we have the transverse velocity is simply going to be the partial derivative of y with respect to t...