- #1
TheExibo
- 55
- 1
Homework Statement
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?
Homework Equations
wavelength=(2(rope length))/nodes
y=Asin(kx)
k=2pi/wavelength
The Attempt at a Solution
[/B]
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?