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Difference between simple harmonic motion and stationary sinusoidal wave?

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tahayassen
#1
Jan24-13, 04:18 PM
P: 273
Their equations are identical. Is there any difference between the two?
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Studiot
#2
Jan24-13, 05:09 PM
P: 5,462
Difficult to coordinate three posts about your lecture notes and experimental lab, but here goes.

No the equations are not the same.

Look again at the equations you posted in your thread about SHM and your notes.

There is something in the equation of a stationary wave that is not in the equation of SHM.

Can you see what it is, because this difference is what makes SHM different from wave motion?
tahayassen
#3
Jan26-13, 10:56 AM
P: 273
According to my professor...

Equation of SHM: [itex]x(t)=Acos(ωt+ϕ)[/itex]

Equation of stationary sinusoidal wave: [itex]x(t)=Acos(ωt)[/itex]

They look almost the same to me. I think she forgot the ϕ in the stationary sinusoidal wave equation.

According to my textbook, the position of SHM is a function of time while the position of a stationary sinusoidal wave is a function of x, but it doesn't actually give me an equation for a stationary sinusoidal wave.

Correct (?) equation of stationary sinusoidal wave: [itex]y(x)=Acos(ωx+ϕ)[/itex]

vanhees71
#4
Jan26-13, 12:03 PM
Sci Advisor
Thanks
P: 2,539
Difference between simple harmonic motion and stationary sinusoidal wave?

What textbook is this? Is this really in there? Then, I'd recommend to choose another one!

The wave equation, say for a string fixed between two points is given by
[tex]\frac{\partial^2}{\partial t^2}u(t,\vec{x})-c^2 \frac{\partial^2}{\partial x^2} u(t,\vec{x})=0,[/tex]
where [itex]u(t,\vec{x})[/itex] is the displacement of the string at position [itex]x[/itex] at time [itex]t[/itex].

Its plane-wave solution reads
[tex]u(t,\vec{x})=u_0 \cos(\omega t \pm k x) \quad \text{with} \quad \omega=c k.[/tex]
Here, [itex]\omega=2 \pi f[/itex] is the "angular frequency" ([itex]f[/itex] is the frequency) and [tex]k=2 \pi/\lambda[/tex] is the "wave number" ([itex]\lambda[/itex] is the wavelength). The upper sign describes a wave running to the left, the lower sign one running to the right.

For the string example, you must also fulfill boundary conditions, describing the fact that the points [itex]x=-L/2[/itex] and [itex]x=+L/2[/itex] are fixed:
[tex]u(t,\pm L/2)=0.[/tex]
Here [itex]L[/itex] is the length of the string.

To find the alowed harmonic solutions you have to superimpose the left- and right-going solutions and determine the coefficients by fulfilling the boundary conditions:
[tex]u(t,x)=A \cos(\omega t-k x)+B \cos(\omega t + k x).[/tex]
To find the coefficients we use the addition theorems for cos:
[tex]u(t,x)=A [\cos(\omega t) \cos(k x)-\sin(\omega t) \sin(k x)] + B [\cos(\omega t) \cos(k x)+\sin(\omega t)\sin(k x)].[/tex]
This can be rearranged to
[tex]u(t,x)=(A+B) \cos(\omega t) \cos(k x) + (B-A) \sin(\omega t) \sin(k x).[/tex]
To fulfill the boundary conditions
[tex]u(t,\pm L/2)[/tex]
one must either have [itex]A=B[/itex] and [itex]k L/2=(2n+1) \pi/2[/itex] or [itex]A=-B[/itex] and [itex]k \lambda/2=n \pi[/itex], where [itex]n \in \mathbb{N}[/itex].

The ground wave is the one with the lowest allowd frequency, corresponding to the lowest allowed [itex]k[/itex], and that's the case 1, i.e., [itex]A=B[/itex] and [itex]n=0[/itex], i.e.,
[itex]k=\pi/L[/itex] and [itex]\omega= c \pi/L[/tex]. The other plane-wave solutions refer to the higher harmonics.

Any motion of the string can be described as a superposition of these socalled "normal modes". As you see from the fact that this same physics describes the sound of such different instruments, based on the motion of strings as a violine, a guitar, or a piano, making pretty different kinds of tones, the wave equation describes quite a large variety of motions of extended bodies (similar wave equations hold for any kind of sound or water waves and many other waves like those of the electromagnetic field, including light).

Compared to this plethora of phenomena the simple harmonic oscillator is quite unexciting. It just describes the motion of a single point-like body attached to a spring or (approximately) a pendulum in the earths gravitational field and the like. It simply oscillates with the one frequency and doesn't do much more than just going back and forth ;-).

The mathematical difference is that the wave equation is a partial differential equation and the simple harmonic oscillator is described by an ordinary differential equation.
Studiot
#5
Jan26-13, 12:41 PM
P: 5,462
I think this material will take a couple or three posts to get through, but there are some underlying ideas that you will be able to take forward to many areas of maths and physics, so it is worth the effort.

First take these ideas through the discussion with you. We can strengthen it later.

SHM is a form of motion that applies to a single particle. It does not propagate in space, although the particle itself moves through a fixed path. Since only one particle is involved there is one single format for SHM. An example is a plumb bob (pendulum) swinging back and forth. The particle is often called an oscillator or a vibrator and the motion an oscillation or a vibration.

A wave is a form of motion that is undergone by an array of particles that are mechanically coupled so that the motion passes from one particle to the next continuously. We call this array the medium of propagation. Since there are many particles involved there is more than one format to wave motion. Examples are the waves on the surface of the sea and the vibrations of the string of a guitar.

Let us look at the equations you have been given and see what can be made of them.

Both wave motion and SHM (and all other forms of motion) are controlled by what is called an equation of motion. This links some quantity to the time and space axes, x and t.

Clasically a single particle can only be in one place at a time and the quantity we choose is position. This leads to your first equation.

x(t) = Acos(ωt+[itex]\varphi[/itex])

This describes the position of the particle, which is oscillating along the x axis, at any instant t.

If you differentiate x with respect to time dx/dt = velocity and you arrive at your second equation.

v(t) = -Aωsin(ωt+[itex]\varphi[/itex])

Given any value of t, this descibes the velocity of the particle at that instant. Remember that the particle is at only one point x at that instant.

Now a wave extends over a region of space. The quantity we choose on the left hand side of the equation of motion will vary from point to point in space in a systematic manner. This quantity is not position (like with SHM). We call it the wave variable. It may be vertical displacement as in the case of the guitar string, or it may be a more complicated property like pressure in a sound wave.

Sticking with the guitar string, the vertical displacement, y is given by your equation for a wave.

Your equation is not correct, which is why I asked if you could see what was missing. It needs to contain the wave variable (Y in the case of our string) on the LHS and the independent variables position (x) and time (t) in the RHS.

Then we can say Y = Y(t,x) ie Y is a function of position and time.

This leads to the equation for a travelling wave

Y(x,t) = Acos (ωt-x)

How are we doing so far?
Do you want to ask questions at this stage or continue to the next stage which is to move on to the stationary wave?


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