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

[Hibiscus]

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.

 

[alphysicist]

HI Hibiscus,

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

<br /> y = A \sin(k x -\omega t)<br />

<br /> y = A \sin[ k (x-vt)]<br />

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

 

[Hibiscus]

HI Hibiscus,

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

<br /> y = A \sin(k x -\omega t)<br />

<br /> y = A \sin[ k (x-vt)]<br />

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.

 

[alphysicist]

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 \omega you can find how v and \omega are related. It's really the same thing either way you solve it.)

 

[Hibiscus]

Ah! Got it! Thank you! =D