What is the wave speed of a transverse wave with given parameters?

  • Thread starter Thread starter Hibiscus
  • Start date Start date
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
Hibiscus
Messages
3
Reaction score
0

Homework Statement


A transverse wave is represented by the function y = 2.3 sin (1.90 x - 25.0t) meters where y, x and t are in meters, meters and seconds respectively. Determine the wave speed in m/s.


Homework Equations


x = A_0 sin (kx - vt)


The Attempt at a Solution


If it's asking for the wave speed, and according to the formula, the wave speed should be 25 m/s and yet the answer is not that. I don't understand what exactly I'm not getting or how I should proceed from here. Any help would be appreciated, thanks.
 
Physics news on Phys.org
HI Hibiscus,

The equation that you have in section 2 does not look right. Here are two forms for the wave equation:

[tex] y = A \sin(k x -\omega t)[/tex]

[tex] y = A \sin[ k (x-vt)][/tex]

The equation they give matches the first form, so 25 is the angular frequency.
 
alphysicist said:
HI Hibiscus,

The equation that you have in section 2 does not look right. Here are two forms for the wave equation:

[tex] y = A \sin(k x -\omega t)[/tex]

[tex] y = A \sin[ k (x-vt)][/tex]

The equation they give matches the first form, so 25 is the angular frequency.

Ah. I guess I put it wrong. In the answer sheet, the form you gave is the way in which it appears. It says, y = A_o sin (k (x - vt) and then it says to equate the terms. And I still don't understand how.
 
So you're given y=2.3 sin (1.90 x - 25.0t)

Rewrite it so it fits the form y = A sin[ k (x - v t) ]

So what do you have to do to (1.9 x - 25.0 t) to get it in the form

k ( x - v t)


(Or since you know [itex]\omega[/itex] you can find how v and [itex]\omega[/itex] are related. It's really the same thing either way you solve it.)