00:01
So it's only asking us to find the amplitude and the speed.
00:03
So we can say that the amplitude, rather, let's find the wavelength first.
00:08
So the wavelength of this transverse is simply two times the length.
00:12
It's going to be two times 0 .386 meters.
00:18
And this will be 0 .772 meters.
00:22
At this point, we can say that the velocity max is going to be equal to the amplitude times omega.
00:29
This is going to be the transverse rather the it will be the velocity not the velocity of propagation don't get confused with this velocity and the velocity of propagation and then the acceleration max would simply be a omega squared we can say that the maximum acceleration divided by the maximum velocity is going to be equal to a omega squared divided by a omega or simply omega and this is going to be equal to 8400 meters per second squared this is the maximum acceleration that they gave us and then they gave us the maximum velocity as well so we can use 3 .80 meters per second and this is going to give us 200 2 ,210 .53 radians per second and at this point we can say okay let's try to solve for a.
01:34
So a can be equal to velocity max over omega.
01:42
So we can say that the a is going to be equal to velocity max, which we already know, 3 .8 meters, 3 .80 meters per second.
01:52
And then this will be divided by the angular frequency of 200 and, rather 2 ,210 .53 radians per second...