00:01
All right, so we've got an equation for simple harmonic motion that says that y of x and t is 0 .105, presumably this is in meters, times the sign of pi x over 9 plus 5 pi times t.
00:20
And part a asks us, what is the transfer speed at t equals 0 .18 seconds? so the speed of the wave, we'll call this v, is just going to be the derivative of this position that we have.
00:37
So with respect to t.
00:40
So this will be 5 pi times 0 .105 meters.
00:46
Sorry, let me write this more clearly.
00:48
5 pi radians per second times 0 .105 meters times the cosine of pi x over 9.
01:02
Plus 5 pi times t and we want to plug in for t sorry t is 0 .18 x is 1 .7 so if we start to plug in our numbers we'll have 1 .7 times pi divided by 9 plus 5 times pi times 0 .18 and then if we take the cosine of that number in radiance multiply by 5 times pi by 0 .105 will get a velocity at this instant of negative 1 .59 meters per second.
01:45
Part b asks, what is the transverse acceleration at the same time? so we can compute the acceleration.
01:53
It turns out it's really just going to be like negative omega squared times y.
01:59
At or sorry x and t this is a function of position and time it's just going to be basically 25 pi squared times y of t and so if we just find what the position is at the particular time specified then we can find out what the acceleration is by multiplying by 25 pi squared so if we do that we should get the acceleration as negative 68 radians per second, or, sorry, meters per second squared.
02:40
And then part c, what is the wavelength of this wave? so we have kx equals pi over nine times x.
02:50
So, of course, our wave number is pi over nine...