Standing waves and tension

1. Apr 21, 2010

zell_D

1. The problem statement, all variables and given/known data
see question 2a, b

2. Relevant equations
a) v=f(wavelength)
b) v= sqrt(F/(m/L))

3. The attempt at a solution
a) f=2.5 Hz, t=1.0 s and one wavelength on the graph is I guess 0.4 m?
v=2.5(.4)=1 m/s
b) f=5.0 Hz, tension same
v= sqrt(F/(m/L)) means that a change in frequency will not change the velocity?
1=5(wavelength)
wavelength= 0.2 m?

1. The problem statement, all variables and given/known data
see question 3a, b

2. Relevant equations
fn=(n/2L)(sqrt(F/(m/L))

3. The attempt at a solution
a)since this is a second harmonic, f2 = 2(2/L)(sqrt(F/(m/L)) = (1/L)(sqrt(F/(m/L))... wavelength = L

quadrupling F would mean that f=(1/L)(sqrt(4F/(m/L)) = (2/L)(sqrt(F/(m/L))
since f does not change, only way that these two will be equal is if the above is divided by 2
(2/2L)(sqrt(F/(m/L)) = (1/L)(sqrt(F/(m/L))
thus wavelength = 2L, so the new picture should just contain half of the wave?

b)tripling tension would mean that wavelength = sqrt(3)L... this shouldn't be possible right? since a standing wave needs to be connected at both ends (nodes) and having a wavelength of 1.7L would not connect on one end...

Last edited: Apr 21, 2010
2. Apr 21, 2010

MalachiK

I think you should start off by looking at the first calculation. You've picked up an extra factor of 10 somewhere.

3. Apr 21, 2010

zell_D

edited, but do things seem right or are they wrong 2b and 3a, b are the ones im not sure on