# Wave Motion, writing an equation.

## 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.

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

Last edited:

Anyone have any idea if I am correct?

alphysicist
Homework Helper
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.

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
Homework Helper
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).

Alright, thank you for the help.

Redbelly98
Staff Emeritus
Homework Helper
$$y(x,t) = A\sin(kx + vt + \phi)]$$

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

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
Staff Emeritus
Your expression (using sin) looks good. 