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2007-03-14 17:46:18 · 2 answers · asked by n a s i m 1 in Science & Mathematics Physics

2 answers

Hmm... does this count?

I believe that simple longitudinal waves can be expressed like springs--however alas I'm not sure! This is assumed for the rest of this answer, but even if I've assumed incorrectly you at least get to see the derivation for an oscillating spring, and a partial derivation of a wave. Perhaps someone else can cover the rest; I'll be keeping an eye on this topic!

Well, we know that a wave is sinusoidal in nature. Thus, a wave can be expressed as either the sine or cosine function, where they differ only by a phase shift, which will later be absorbed into φ. (Recall sin(θ) = cos(π/2 - θ).)

Now, our displacement (x) as a function of time (t) will be expressed as:
x(t) = A*sin(θ), where A is the amplitude of the wave, and θ is a function of t.

We can express θ(t) = ωt + φ, where ω is our angular frequency in radians per second (notice this works by dimensional analysis, where [rad/s]*[s] = [rad]), and φ is our phase shift in radians.

Thus:
x(t) = A*sin(ωt + φ)
Or, for that matter:
x(t) = A*cos(ωt + φ)
Where φ is a constant phase shift which will depend on the problem, and which of the two aforementioned conventions (expressed as sine or cosine) is used.

Likewise, velocity and acceleration as a function of time can be found by taking the first and second derivatives, respectively.

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With a bit of differential equations (D.E.) you can derive it from "scratch."

For the sake of simplicity, we'll consider a simple spring, where by Hooke's Law: F=-k*x, where k is the spring constant in Newtons per meter, and x is the displacement of the spring (or any oscillatory device, for that matter) from equilibrium expressed in meters.

Since, by Newton's Second Law, F=m*a, and a is just the second derivative of x, we have a D.E. where:
m*x'' + k*x = 0
We can simplify this:
x'' + (k/m)*x = 0
Which we can solve using the Method of Undetermined Coefficients to be:
x(t) = c1*cos(ωt) + c2*sin(ωt)
Where c1 and c2 are initial conditions, we'll assign them:
c1 = A*cos(φ)
c2 = A*sin(φ)
Where φ is a constant representing the phase shift, and A a constant representing amplitude. Thus:
x(t) = A*cos(φ)*cos(ωt) + A*sin(φ)*sin(ωt)
Which by our Addition Formula for Cosine gives us:
x(t) = A*cos(ωt - φ)
But since φ is just a constant, we can declare it positive or negative in our general form, thus it can be rewritten one final time:
x(t) = A*cos(ωt + φ)
(And likewise we could replace cosine with sine, as mentioned way above.)

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Similar, but more complex techniques can be used to derive the equation of a wave that is damped (as most oscillations are in real life), or driven. Moreover, this represents the equation of a longitudinal wave, whereas most waves in "real life" are not so ideal...

For reference, a transverse wave is expressed:
y(x,t) = A*sin(kx ± ωt + φ)
Or
y(x,t) = A*cos(kx ± ωt + φ)

Where the y position depends also on the x position. Alas, I'm unsure how to derive this.

I hope this helps!

2007-03-14 18:01:16 · answer #1 · answered by Brian 3 · 0 0

Be more specific please..

What kind of wave?

2007-03-15 00:54:26 · answer #2 · answered by Boozer 4 · 0 0

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