Wave Motion, writing an equation.

[Ithryndil]

Homework Statement

A sinusoidal wave traveling in the -x direction (to the left) has an amplitude of 20.0 cm, a wavelength of 29.0 cm, and a frequency of 15.0 Hz. The transverse position of an element of the medium at t = 0, x = 0 is y = -3.00 cm, and the element has a positive velocity here.

Write an expression for the wave function y(x,t), where y and x are expressed in cm, and t is expressed in seconds.

Homework Equations

$$\omega = 2\pif$$

v=(lambda)(frequency)

$$k = 2\pi/\lambda$$

$$y(x,t) = Asin[\left(2\pi/\lambda\right)(x - vt)]$$

The Attempt at a Solution

We are given the fact that:

$$A = 20.0cm$$

$$\lambda = 29cm$$

$$v = (lambda)(frequency) = 435cm/s$$

I think the answer is the following, but I am not sure, and I only have one more submission left, so I want to be sure.

$$y(x,t) = -20.0cos[.217(x-435t)]$$

 

[Ithryndil]

Anyone have any idea if I am correct?

 

[alphysicist]

Hi Ithryndil,

If you check your answer at x=0 and t=0, you get that y=-20 cm. However, this is not what the problem specifies; they say they want x==-3cm at x=0 and t=0.

I think you need to include a phase constant in your expression that will give the correct initial condition.

 

[Ithryndil]

You mean y = 3 cm? And I think so too. I just realized that the general equation for y(x,t) is as follows:

$$y(x,t) = Asin(kx - vt + \phi)]$$

I think the above equation is for a wave traveling to the right. I think because this wave is traveling to the left it needs to be:

$$y(x,t) = Asin(kx + vt + \phi)]$$

 

[alphysicist]

Yes, I did mean y= -3 cm. Sorry!

Your updated equation looks good. Once you plug in the numbers you found, you can check that it matches y=-3cm at x=0 and t=0 and that it also has a positive velocity (in the y direction at that point).

 

[Ithryndil]

Alright, thank you for the help.

 

[Redbelly98]

$$y(x,t) = A\sin(kx + vt + \phi)]$$

That is almost correct. There should either be some more parentheses inside the sin argument, or replace vt with $$\omega t$$.

 

[Ithryndil]

You're right, it should be: $$y(x,t) = A\sin(kx + \omegat + \phi)$$

I wound up getting the following for the equation:

$$y(x,t)20.0sin(0.217x+94.25t-0.1506)$$

Now I was stupid and input $$y(x,t)20.0cos(0.217x+94.25t-0.1506)$$...note I put cosine and not sine. Yes, the cosine would be correct with the right phase angle, but it's not with that phase angle.

 

[Redbelly98]

Your expression (using sin) looks good.

 

[Ithryndil]

Thank you. I just can't believe I inputted the expression with cos instead of sin.