1. Continua Continued More Useful: find Normal Mode solutions:
good solution if…
therefore if…
… normal modes have a sinusoidal shape.

2. Set f = p/2 for the moment and apply product-to-sum:
Rearrange to f(x-vt)
Normal mode = two counter-propagating traveling waves

3. node
antinode

4. L, m, T Infinite strings give infinite solutions. Try a bound string:
f(0)=0 and f(L)=0:
L, m, T
Wave equation still applies.
Normal mode solutions still apply.
But now we can apply boundary conditions
Up to now we have only talked about harmonic oscillators.
We have done it quickly and not focused on every detail because the class is supposed to be about Waves!
Also, our study on oscillators showed us many useful math techniques and physical concepts.
A wave is the propagation of energy through a medium, a oscillator is not a medium!
However, two coupled oscillators is getting there: saw energy transfer last time.
Many coupled oscillators starts to look like a medium, but this is the time to switch from coupled oscillators to an elastic continuum.
This stretched string (above) is the standard continuum when learning physics: if you divide it into smaller and smaller pieces, it still looks like string.
Here is a real stretched string: the idealization breaks down on many levels: at a very small scale its made of atoms – not a continuum! At a larger scale its really nylon surrounding a rubber core, don’t have to magnify much to see that!
If I pluck it, you can see that it supports wave propagation, so it should serve us well.
We characterize the string with a length (here its 4.4 m), a mass/length (here its kg/m), and a Tension (here this weight makes it 4.2 N)
To describe its motion, we consider a coordinate system with x along the length of the string, and displacements of the string in y. So we get a function y(x,t) that describes transverse displacement as a function of position and time. So we are looking at transverse waves. Longitudinal waves are also possible, but harder to see, and they don’t really help you mathematically.
This function y(x,t) is going to describe the motion just like x=Asin(wt+p) did for the oscillators. Therefore, we will need to use derivatives, and when you have a function of two variables you need partial derivatives. These are very similar to derivatives you are used to…
Boundary conditions reveal specific normal mode frequencies!
(for the clamped string)

6. These are the normal mode solutions, NOT the solution to some driving force or initial condition!!

7. If we shake one end….
With these Boundary conditions: y L x
… is this a good solution?
Boundary conditions reveal normal mode wavelengths & frequencies!
(for the clamped string)

8. Apply boundary condition:
Sine is zero if:
Apply other boundary condition:

9. n = 1
B = 0.1
L = 10
v = 1
w = 1.05 (npv/L)

10. n = 2
B = 0.1
L = 10
v = 1
w = 1.05 (npv/L)

11. n = 5
B = 0.1
L = 10
v = 1
w = 1.05 (npv/L)

12. n = 5
B = 0.1
L = 10
v = 1
w = 1.01 (npv/L)

13. n = 5
B = 0.1
L = 10
v = 1
w = (npv/L)

14. Wavenumber:
French is the only book still in print that defines k = 1/l !!!
The equation of motion for a stretched string (a continuum) is the wave equation. It has traveling wave and normal mode solutions (which are really the same thing).