# What is the value of ω for the following wave?

Eitan Levy
Homework Statement:
There is an electromagnetic wave with a magnetic component equal to:

Bsin(ax+by-3*sqrt(5)*10^6*t)(2x/3-y/3-2z/3) (The second set of parentheses represent the direction of the wave)

What is ω for this wave?
Relevant Equations:
ω=kv
I am quite confused by this. I was sure that ω=3*sqrt(5)*10^6, because that is the coefficient of t, and generally u(x,y,z,t)=Asin(kx−ωt+ϕ) for a multi dimensional harmonic wave.

However in the answers it is said that ω=3*sqrt(5)*10^14. I can't see the reason for that, could anyone explain please?

Homework Helper
Gold Member
What are the units in the answer that is given to you? What are the units in the answer that you think is correct? Do you have numbers and units for a and b?

Eitan Levy
What are the units in the answer that is given to you? What are the units in the answer that you think is correct? Do you have numbers and units for a and b?
No units were provided. The question is exactly what I wrote.

Mentor
Bsin(ax+by-3*sqrt(5)*10^6*t)(2x/3-y/3-2z/3) (The second set of parentheses represent the direction of the wave)
So using the LaTeX Guide at the bottom of the edit window, is this your expression?

$$B sin(ax + by - 3\sqrt5 10^6 t) (\frac{2}{3}\hat x - \frac{1}{3} \hat y - \frac{2}{3} \hat z)$$

If so, it seems strange that the argument to the sin() function is dependent on x and y and t. But maybe I'm missing something... Maybe the medium that the wave is propagating though is non-isotropic?

Eitan Levy
So using the LaTeX Guide at the bottom of the edit window, is this your expression?

$$B sin(ax + by - 3\sqrt5 10^6 t) (\frac{2}{3}\hat x - \frac{1}{3} \hat y - \frac{2}{3} \hat z)$$

If so, it seems strange that the argument to the sin() function is dependent on x and y and t. But maybe I'm missing something... Maybe the medium that the wave is propagating though is non-isotropic?

It is and there was a mistake in the answers.

• berkeman
$$B sin(ax + by - 3\sqrt5 10^6 t) (\frac{2}{3}\hat x - \frac{1}{3} \hat y - \frac{2}{3} \hat z)$$
The expression is probably a special case of $$\vec B(\vec r,t)=\vec B_0\sin(\vec k\cdot\vec r-\omega t)$$with ##k_x=a## and ##k_y=b##. Of course the values of ##a## and ##b## must be adjusted to ensure that the given direction of the field is orthogonal to the direction of propagation.
• 