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How Are k Omega And v Derived

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How Are k Omega And v Derived

Question

How does the board derive the relations among k, omega, lambda, T, f, and wave speed v?

Starting Point

A traveling wave can be written as:

y(x,t) = A cos(kx - omega t + phi)

Where:

  • k is the wave number
  • omega is the angular frequency
  • phi is the phase constant

The key idea is:

  • if the phase changes by 2pi, the physical wave is unchanged

Deriving k = 2pi / lambda

Compare two points separated by one wavelength lambda:

y(x,t) = A cos(kx - omega t + phi)
y(x + lambda, t) = A cos(k(x + lambda) - omega t + phi)
                  = A cos(kx - omega t + phi + k lambda)

Because points separated by one wavelength have the same wave value, the phase change must be 2pi:

k lambda = 2pi

So:

k = 2pi / lambda

Deriving omega = 2pi / T = 2pi f

Now compare times separated by one period T:

y(x, t + T) = A cos(kx - omega(t + T) + phi)
            = A cos(kx - omega t + phi - omega T)

After one period, the wave repeats, so the phase change is again 2pi:

omega T = 2pi

Thus:

omega = 2pi / T

Since:

f = 1 / T

we get:

omega = 2pi f

Deriving v = omega / k

A moving crest keeps constant phase:

kx - omega t + phi = constant

Rearrange:

kx - omega t = constant
=> x = (omega / k)t + constant

So the propagation speed is:

v = omega / k

Recovering v = f lambda

Substitute:

omega = 2pi f
k = 2pi / lambda

Then:

v = (2pi f) / (2pi / lambda) = f lambda

Final Relations

The board is deriving these standard wave relations:

k = 2pi / lambda
omega = 2pi f
v = omega / k = f lambda

Counterpoints and Gaps

  • this derivation is for a standard traveling wave form and does not yet discuss sign conventions such as kx + omega t
  • it also does not yet distinguish phase velocity from group velocity

Graph View

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