Waves and Optics by Paul Padley - HTML preview

PLEASE NOTE: This is an HTML preview only and some elements such as links or page numbers may be incorrect.
Download the book in PDF, ePub, Kindle for a complete version.

Chapter 3Mechanical Waves

3.1Vibrations on a String*

Vibrations on a String

Figure (String-Fragment-small.gif)
Figure 3.1

Consider the forces on a short fragment of string Fy = T sin ( θ + Δ θ ) − T sinθ Fx = T cos ( θ + Δ θ ) − T cosθ Assume that the displacement in y is small and T is a constant along the stringthus θ and θ + Δ θ are smallthen Fx ≈ 0 we can see this by expanding the trig functions _autogen-svg2png-0007.png or FxT θ Δ θ which is very small.On the other hand FyT [ θ + Δ θθ + ] or FyT Δ θ which is not nearly as small. So we will consider the y component of motion, but approximate there is no x component _autogen-svg2png-0012.png Also we can write: Fy = m ay m = μ Δ x where μ is the mass density _autogen-svg2png-0016.png now have _autogen-svg2png-0017.png_autogen-svg2png-0018.png Note dimensions, get a velocity _autogen-svg2png-0019.png _autogen-svg2png-0020.png The second space derivative of a function is equal to the second time derivative of a function multiplied by a constant.

Normal Modes on a String

Before considering traveling waves, we are going to look at a special case solution to the wave equation. This is the case of stationary vibrations of a string.

For example here, lets consider the case where both ends of the string are fixed at y = 0 . Now we vibrate the string. Every point along the string acts like a little driven oscillator. So lets assume that every point on string has a time dependence of the form cos ωt and that the amplitude is a function of distance Assume y ( x , t ) = f ( x ) cos ωt then _autogen-svg2png-0024.png _autogen-svg2png-0025.png Substitute into wave equation _autogen-svg2png-0026.png _autogen-svg2png-0027.png Then every f ( x ) that satisfies: _autogen-svg2png-0029.png is a solution of the wave equation

A solution is (requiring f ( 0 ) = 0 since ends fixed) _autogen-svg2png-0031.png Another boundary condition is f ( L ) = 0 so get _autogen-svg2png-0033.png Thus _autogen-svg2png-0034.png _autogen-svg2png-0035.png

Be careful with the equations above: v is the letter vee and is for velocity. now we introduce the frequency ν which is the Greek letter nu.

recall ν = ω / 2 π so _autogen-svg2png-0039.png This is a very important feature of wave phenomena. Things can be quantized. This is why a musical instrument will play specific notes. Note, that we must have an integral number of half sine waves _autogen-svg2png-0040.png end up with _autogen-svg2png-0041.png leading to _autogen-svg2png-0042.png where _autogen-svg2png-0043.png ω1 is the fundamental frequency

3.2Waves*

Waves

The Wave Equation

In deriving the motion of a string under tension we came up with an equation:

_autogen-svg2png-0001.png which is known as the wave equation. We will show that this leads to waves below, but first, let us note the fact that solutions of this equation can be added to give additional solutions.

Waves Add

Say you have two waves governed by two equations Since they are traveling in the same medium, v is the same_autogen-svg2png-0003.png_autogen-svg2png-0004.pngadd these_autogen-svg2png-0005.png_autogen-svg2png-0006.pngThus f1 + f2 is a solution to the wave equation

Lets say we have two functions, f1 ( x − v t ) and f2 ( x + v t ) . Each of these functions individually satisfy the wave equation. note that y = f1 ( x − v t ) + f2 ( x + v t ) will also satisfy the wave equation. In fact any number of functions of the form f ( x − v t ) or f ( x + v t ) can be added together and will satisfy the wave equation. This is a very profound property of waves. For example it will allow us to describe a very complex wave form, as the summation of simpler wave forms. The fact that waves add is a consequence of the fact that the wave equation _autogen-svg2png-0013.pngis linear, that is f and its derivatives only appear to first order. Thus any linear combination of solutions of the equation is itself a solution to the equation.

General Form

Any well behaved (ie. no discontinuities, differentiable) function of the form y = f ( x − v t ) is a solution to the wave equation. Lets define _autogen-svg2png-0016.pngand _autogen-svg2png-0017.pngThen using the chain rule _autogen-svg2png-0018.pngand _autogen-svg2png-0019.pngAlso _autogen-svg2png-0020.png_autogen-svg2png-0021.pngWe see that this satisfies the wave equation.

Lets take the example of a Gaussian pulse. f ( x − v t ) = A e − ( x − v t ) 2 / 2 σ2

Then _autogen-svg2png-0023.png

and _autogen-svg2png-0025.png or _autogen-svg2png-0027.png That is it satisfies the wave equation.

The velocity of a Wave

Figure (Traveling-Wave-small.png)
Figure 3.2

To find the velocity of a wave, consider the wave: y ( x , t ) = f ( x − v t ) Then can see that if you increase time and x by Δ t and Δ x for a point on the traveling wave of constant amplitude f ( x − v t ) = f ( ( x + Δ x ) − v ( t + Δ t ) ) . Which is true if Δ x − v Δ t = 0 or _autogen-svg2png-0033.png Thus f ( x − v t ) describes a wave that is moving in the positive x direction. Likewise f ( x + v t ) describes a wave moving in the negative x direction.

Lots of students get this backwards so watch out!

Another way to picture this is to consider a one dimensional wave pulse of arbitrary shape, described by y = f ( x ) , fixed to a coordinate system O ( x , y )

Figure (IKYTDO01.png)
Figure 3.3

Now let the O system, together with the pulse, move to the right along the x-axis at uniform speed v relative to a fixed coordinate system O ( x , y ) .

Figure (IKYTFW02.png)
Figure 3.4

As it moves, the pulse is assumed to maintain its shape. Any point P on the pulse can be described by either of two coordinates x or x, where x = x − v