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Antinodes and Nodes on a Waveworm

tristancohn

New member
Sep 12, 2013
2
i have a waveform:
y2(x)= sin(x*1.2)+sin(x*1.8)

the first 6 nodes, meaning the parts of the wave where y2=0 come up on my graph at around x =

2.094
4.189
5.236
6.283
8.377
10.471

you can find the nodes easily (where y2=0) with either x = n*pi/(1.5) or x = (2n+1)*(pi/0.6), where n = 0, 1,2 3, and so on.

what I really want to find is the antinodes (peaks) of y2, similar to on this graph where the line turns black:

http://upload.wikimedia.org/wikipedia/commons/7/7a/Graph_of_sliding_derivative_line.gif

which are at around x =

1.006
2.979
4.665
5.808
7.492
9.466

I know that the antinodes for y2 come up at 1.8*COS(1.8*x)+1.2*COS(1.2*x)=0
but i just can't find a clean way of finding them.
:confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We are given the waveform:

\(\displaystyle y_2(x)=\sin\left(\frac{6}{5}x \right)+\sin\left(\frac{9}{5}x \right)\)

It appears what you have done to find the nodes is apply a sum to product identity to get:

\(\displaystyle y_2(x)=2\sin\left(\frac{3}{2}x \right)\cos\left(\frac{3}{10}x \right)\)

Because, in this form, the angular velocity of the sine factor is an odd multiple of that of the cosine factor, the period $T$ of the function is therefore:

\(\displaystyle T=\frac{10}{3}\pi\)

and the roots (nodes) are at:

\(\displaystyle x=\frac{2k}{3}\pi,\,\frac{5}{3}(2k+1)\pi\) where \(\displaystyle k\in\mathbb{Z}\)

Here is a plot of the function over one period:

tristancohn.jpg

The nodes for this period are at:

\(\displaystyle x=0,\,\frac{2}{3}\pi,\,\frac{4}{3}\pi,\,\frac{5}{3}\pi,\,2\pi,\,\frac{8}{3}\pi,\,\frac{10}{3}\pi,\,\)

So far , so good. To find the anti-nodes, we obviously want to equate the derivative with respect to $x$ of the given waveform to zero:

\(\displaystyle y_2'(x)=\frac{6}{5}\cos\left(\frac{6}{5}x \right)+\frac{9}{5}\cos\left(\frac{9}{5}x \right)=0\)

Multiplying through by \(\displaystyle \frac{5}{3}\) and letting \(\displaystyle u=\frac{3}{5}x\) we obtain:

\(\displaystyle 2\cos\left(2u \right)+3\cos\left(3u \right)=0\)

At the moment, I can't think of a nice clean way to solve this trigonometric equation, but perhaps someone else will know of a method.

Using double and triple angle identities for cosine and letting $v=\cos(u)$, we then get:

\(\displaystyle 12v^3+4v^2-9v-2=0\)

We could obtain exact values for the roots of this cubic, but I find such methods cumbersome, and then we are left to take the inverse cosine function of these roots anyway, so I suggest we approximate the roots.

Using a numeric root finding method, we find:

\(\displaystyle v\approx-0.941785319480139,\,-0.214929385113036,\,0.823381371259841\)

Back-substituting, where \(\displaystyle x=\frac{5}{3}\cos^{-1}(v)\), we then find:

\(\displaystyle x\approx1.00575387440944,\,2.979026418974480,\,4.66449632903117\)

By the symmetry of the wave form about \(\displaystyle x=\frac{5}{3}\pi\), we may then give the other anti-nodes in the plotted period as:

\(\displaystyle x\approx5.80747918293481,\,7.492949092991497,\,9.466221637556537\)
 

tristancohn

New member
Sep 12, 2013
2
Back-substituting, where \(\displaystyle x=\frac{5}{3}\cos^{-1}(v)\), we then find:

\(\displaystyle x\approx1.00575387440944,\,2.979026418974480,\,4.66449632903117\)

By the symmetry of the wave form about \(\displaystyle x=\frac{5}{3}\pi\), we may then give the other anti-nodes in the plotted period as:

\(\displaystyle x\approx5.80747918293481,\,7.492949092991497,\,9.466221637556537\)
Thanks, I don't really understand how you back-substitute x though.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I took the approximations for $v$ and used \(\displaystyle x=\frac{5}{3}\cos^{-1}(v)\).