# Displacement of a point on a string given t and x

1. Oct 22, 2008

### beccaka2003

1. The problem statement, all variables and given/known data

The equation describing a transverse wave on a string is
y(x,t) = (3.50mm)sin[(t/162s) - (x/42.5)]

Find the transverse displacement of a point on the string when t = 0.220 s and at a position x = 0.140m

2. Relevant equations
y(x,t) = Asin(wt-kx) or y(x,t) = Asin2pi[(t/T)-(x/wavelength)]
k = 2pi/wavelength
w = 2pif

3. The attempt at a solution

I know from previous parts of the problem that I got right:

wavelength = 0.148m
f = 1/T = 25.8 Hz
A = 3.5mm

I thought that you would just plug the values of t and x into the second equation I gave, but it was wrong. I keep on coming up with y = 1.73, no matter what I do, but the answer is y = -3,46. What am I doing wrong?

2. Oct 23, 2008

### alphysicist

Hi beccaka2003,

This is the equation that you want to plug your t and x value into. But isn't this equation wrong? I have found a form that matches the frequency and wavelength you said you got on the previous parts of the problem, and it does not match the above formula.

So please check and fix this equation. Do you get the right answer?

3. Oct 23, 2008

### beccaka2003

No, that is the equation that we are given. In the homework, it was written slightly different, but I'm positive they are the same:

y(x,t) = (3.55mm)sin[(162s^-1)t-(42.5m^-1)x]

They are the same, right? I got the correct answers for the first 3 parts of the question using it.

I really want to understand this, and it is driving me nuts.

Last edited: Oct 23, 2008
4. Oct 23, 2008

### alphysicist

From your wavelength and frequency values:

wavelength = 0.148m
f = 1/T = 25.8 Hz

I'm getting that the w and k values are:

k=42.5 m-1

If you then look at your general equation:

y(x,t) = Asin(wt-kx)

you can see that t is multiplied by w, not divided; and x is multiplied by k, not divided. So I think your original equation should not be:

y(x,t) = (3.50mm)sin[(t/162s) - (x/42.5)]

it should be:

y(x,t) = (3.50mm)sin[(162) t - (42.5) x]

Are you sure your book did not have it in this form? I'm confused as to how you could have gotten the correct wavelength and frequency if it was not in this form.

Do you get the right answer?

5. Oct 23, 2008

### alphysicist

(I was responding to your post while you were still modifying it, so you might not have seen my last post.)

No, they are not the same. The s-1 and the m-1 are units; the numbers 162 and 42.5 stay in the numerator.

6. Oct 23, 2008

### beccaka2003

Ahh! Its all because my calculator was in degrees, not radians. The reason the other equation worked for me was because I had taken the long way around and made the equation y(x,t) = (3.50mm)sin2pi[(t/25.8s) - (x/.148)]. That should have told me right there that y(x,t) = (3.55mm)sin[(162s^-1)t-(42.5m^-1)x] is not equivalent to
y(x,t) = (3.50mm)sin[(t/162s) - (x/42.5)]. Silly mistake that made a really simple problem impossible. Thanks so much!

7. Oct 23, 2008