Can the angular wave number(k) or frequency(w) be negative?

[Ethan Godden]

1.Homework Statement

The wave function for a wave on a taunt string is:

y(x,t)=(0.350)(sin(10(π)(t)-3(pi)(x) +(π)/4)

where x and y are in meters and t is in seconds. If the linear mass density(μ) of the string is 75.0g/m, (a) what is tha average rate at which energy is transmitted along the string(P)? (b) What is the energy contained in each cycle of the wave?

Homework Equations

General Wave Function: (A)(sin(kx-ωt))
P=(1/2)(μ)(ω²)(A²)(v)
E_λ=(1/2)(μ)(ω²)(A²)(λ)
k=(2π)/λ
ω=(2π)f
v=λf

The Attempt at a Solution

My main question has to do with whether the angular wave number and/or the angular frequency can be negative. I am comparing the general wave function given in my textbook and the function given in the question, and the variables have different signs such as the "kx" be positive in the general function and negative in the question or the "ωt" being negative in the general function and positive in the question.

My first thought was that maybe the wave in this question is moving to the left as the general wave function changes to "(A)(sin(kx+ωt))" if this is the case, but when I did a bit of searching online, people said the wave is moving to the right. I am perplexed to how one can determine this.

I am pretty sure that whether k and ω are negative for this question does not matter because both would have the same sign as my solution below shows. I am just interested to know why k and ω are not negative and how one could determine the direction of motion for the wave from the quesiton.

Solution:

(a) k=(2π)/λ →λ=(2π/k) →λ=(2π)/(±3π)=±0.6666...rad/m
ω=(2π)f → ω/(2π)=f → ±(10π)/(2π)=±5Hz
v=λf → v=(±5)(±0.6666...)=3.33333...m/s

Input these values into the equation for P to get the average rate at which the energy is transmitted.

(b) Input the values into the equation for E_λ to find the energy contained in each cycle.

Also, if you wouldn't mind, please tell me if I am using this site correctly as this is my first post.

Thank You

 

[TSny]

Welcome to PF!
You did a great job using the template in posting your question.

$k$ and $\omega$ are both positive numbers.

Note that $\sin \left( kx-\omega t \right) = \sin \left(-(\omega t -kx) \right) = -\sin \left(\omega t -kx \right) = \sin \left(\omega t -kx +\pi \right)$

So, $\sin \left( kx-\omega t \right)$ represents the same wave as $\sin \left( \omega t - kx \right)$ except for a difference of $\pi$ in the phase angle. Both ways of writing it represent a wave traveling toward positive x.

If $kx$ and $\omega t$ have opposite sign, the wave travels toward positive x.
If $kx$ and $\omega t$ have the same sign, the wave travels toward negative x.

 

[Ethan Godden]

Thank you for you response,

I have a follow-up question about this.

Since ω=(2π)f and k=(2π)/λ, does this mean f and λ can be negative?

Thank you,

 

[TSny]

No. Since k and ω are positive, so are f and λ. All of these are positive "by definition".

 

[Ethan Godden]

Okay,

To clarify, the positive and negative signs in the function, whatever ever way we chose to write the function, just signal direction and have no association with the variables such as k and ω because they are by definition positive?

 

[TSny]

Okay,

To clarify, the positive and negative signs in the function, whatever ever way we chose to write the function, just signal direction and have no association with the variables such as k and ω because they are by definition positive?

Yes, that's right.

If you go on to study waves propagating in two or three dimension of space, then you will find that k can become a vector quantity. Its x, y, or z components can then be negative. But that's another story. Even in that case, the quantities λ, f, and ω remain positive by definition.