Advanced Math/Calculating the position of the nodes on a waveform
I have a waveform of y1(x)= 1.2*cos(1.2*x)+1.8*cos(1.8*x), and i need to find the position of the nodes.
I found the nodes for y2(x)= sin(x*1.2)+sin(x*1.8) with either x = nπ/(1.5) or x = (2n+1)*(π/0.6), where n = 0, 1,2 3, and so on.
but i can't find an easy way for the first one.
Tristan, it is indeed harder to find nodes for y1(x) than for y2(x). I didn't get a closed form solution but the maybe the following will help.
y1(x) = 1.2cos(1.2x) + 1.8cos(1.8x). The difference in the amplitudes of the cosines is of course what makes it tricky.
We want to find x such that y1(x) = 0 or, letting a = 1.2 and b = 1.8 for now
acos(ax) = -bcos(bx).
My approach is to introduce an increment d to the angle ax and to introduce a factor of π' = (2n+1)π to make the cosine negative so that
bcos(bx) = bcos(ax+d+π') which gives (it helps to draw 2 unit circles of radius a and b and plot these quantities)
d = (b-a)x-π',
which is fine except now we need another independent equation relating d and x to solve the problem. Again, using the unit circles,
acos(ax) = bcos(ax+d) <- no π this time.
Using a trig identity, get
tan(ax) = sin(d)/(cos(d)-a/b)
Substituting the expression for d gives a complicated expression involving x, call it f(x). I don't think it can be solved in closed form for x (it transcendental) so the next step would be to get an approximate value for x by cross plotting
tan(ax) = f(x)
and seeing at what value of x they cross. This should give you the node. Sorry I can't give you a cleaner answer. Let me know if you want me to work some more on this. Good luck.