# Wavelength from amplitude information?

1. Mar 26, 2016

### TheExibo

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

A string vibrates at its third-harmonic frequency. The displacement at a point 59.0cm from one end is the first time the displacement is half the maximum amplitude. How long is the string?

2. Relevant equations

wavelength=(2(rope length))/nodes

y=Asin(kx)

k=2pi/wavelength

3. The attempt at a solution

0.5A=Asin(kx)
kx=30

k(0.59m)=30
k=50.85

50.85=2pi/wavelength
wavelength=0.1236m

0.1236m=(2(rope length))/3
rope length=0.1854m

How is such a small rope length possible if 0.59m is half the distance from the first amplitude? What am I doing wrong?

2. Mar 26, 2016

### Simon Bridge

You have noticed that you have a point on the string that is beyond the length of the string ... this is a contradiction: therefore you did something wrong.
This could be arithmetic, maybe you had the calculator in DEG when it should be in RAD or other way?
It could be something to do with the way you reasoned your way through the problem: cannot tell because you don't say... but I did notice that you did your angles in degrees when you should have used radians.
The angles in all scientific formulas are in radians ...

Note: Best practise is to do all the algebra before you plug the numbers in.

Last edited: Mar 26, 2016
3. Mar 26, 2016

### TheExibo

I used radians, and got a very long length for the wire. I only have one try left for answering my question, so I'm afraid to risk it. My reasoning was to find the wave number using y=Asin(kx) (this equation was suggested in another question, but I don't know if it is right since I've seen y=2Asin(kx) being used), then to use the wave number to find the wavelength, and use the resonance for closed-closed structures to find the length of the rope.

4. Mar 27, 2016

### TheExibo

I found that y=Asin(kx) should be the correct one to use, and with radians, I get 10.62m. I'm still not sure

5. Mar 27, 2016

### TheExibo

Never mind, the answer with the radians was correct. How come degrees wouldnt work though?