00:01
So in question 83, we have the new fluid wave of amplitude yn, and it has a wave lens lambda, and this wave is traveling on a chord.
00:10
So i'm going to find out what is the maximum particle speed, that is to say the transfer speed, and then what is that speed with respect to the wave speed.
00:21
So let's do this first write out the equation of this wave, y equals y n, sine, kx, i think it mentioned negative traveling along the positive direction, so it is minus omega -t.
00:37
And in order to find the transfer velocity or particle velocity, we need to take the derivative of this y.
00:45
So, d -y over d -t equals negative omega -by -n, sine, cosine, kx minus omega -t.
00:57
So yeah, this is a cosine because that is the derivative of sine.
01:00
And okay, so what is the maximum velocity? so maximum velocity, so this is u, and the maximum velocity u max is when cosine is negative 1.
01:17
So this is omega times ym.
01:22
What about the wave speed then? so wave speed v is given by b equals waveless over the time to travel wavelengths.
01:33
And another way of writing it is omega over.
01:40
So part a asks us to find the ratio of it.
01:44
So we are looking for u max over v...