Tension in string and oscillation

1. Mar 5, 2015

gasar8

We've got a 0,5m string attached to a frame and has its own fundamental frequency at 440Hz. We cool our system for 15°C. What is the proper frequency now?

String length l=0,5m
String section S=0,02 mm^2
String density ρ=7800 kg/m^3
Young module E=2*10^5 N/mm^2
Fundamental frequency ν=440Hz
ΔT= -15°C
α(string)=1,2*10^-5 /K
α(frame)=1,7*10^-5 /K

2. Relevant equations
Δl/l=F/(ES)+α⋅ΔT
I assume that F(string)=F(frame) and Δl(string)+Δl(frame)=0, so:

F/(ES) +α(string) ΔT + α(frame) ΔT=0

From this equation I get F=1,74N, but I can't imagine what this number means?

c=√(E/ρ) = 5063 m/s is this equal to √(F/(ρS))? What is this force? (4000N)

2. Mar 5, 2015

Orodruin

Staff Emeritus
Do you know how wave velocity depends on string tension?

3. Mar 5, 2015

gasar8

From this formula √(F/(ρS)) I would say that greater tension produces higher wave velocities.

4. Mar 5, 2015

haruspex

It's hard to be sure from such a skeletal description, but I suspect you have some errors in that calculation.
For a start, I would have thought Δl(string)=Δl(frame).

5. Mar 5, 2015

gasar8

Half a meter long steel string with a cross-section 0,02 mm2 with a density of 7800 kg /m3 and Young module 2*105 N/mm2 is embedded in the massive brass frame and has its fundamental frequency at 440 Hz. What is the new frequency, when the string and the frame is cooled for ΔT= 15°C?
Temperature coefficient of length expansion are αs = 1.2*10-5 / K for the string and αf = 1.7*10-5 / K for the frame.

So:
Δls/l = F/(ES)+αs ΔT
Δlf/l = αf ΔT

Δls = Δlf
F/(ES)+αs ΔT= αf ΔT
F=(αf - αs) ΔT E S
F= -0,3N --> What does this mean?

6. Mar 5, 2015

haruspex

If I rewrite your equation as F/(ES)= αf ΔT-αs ΔT, the right hand side is the change in the stretch of the wire. Does that help?
(I agree with -0,3N.)

7. Mar 8, 2015

gasar8

So if I understand this correctly, it means, that there is -0,3N less force in the string than at the beginning?

For the force at the beginning I use the following formula:
f1=√(F1/(ρS))/2l, where f =440Hz
and get F1=120,8N

Is it correct now, that I use F2=F1-0,3N=120,5N for the final force?
...and get f2=√(F2/(ρS))/2l2=879Hz?
Do I even have to take into account that l2=l-(αf ΔT-αs ΔT), because it is negligibly small?
This result is an octave higher which seems a little bit exaggerated for me at only ΔT=15°C? :)

8. Mar 8, 2015

haruspex

That's double what it should be. Did you take l as 1 instead of .5?
It certainly shouldn't be much changed.
The change in length is not (αf ΔT-αs ΔT). But you are right, it does not have much effect.

9. Mar 9, 2015

gasar8

Ups, yes I took l=1m but if I use the correct value I get 30,2N which is 4 times smaller?

So my thought that new force is 30,2N-0,3N=29,9N is right?
And I get final frequency f2=437 Hz?

Last edited: Mar 9, 2015
10. Mar 9, 2015

haruspex

Sorry for the delay, I went looking for my scribbled answer but couldn't find it. I think that's what I got.

11. Mar 9, 2015

gasar8

Ok, thank you very much for your time and help. :)

12. Mar 11, 2015

gasar8

Hi, I've got one more question for the same exercise:
How does the energy of oscillation change, if the amplitude is 2mm at the beginning and at the end if the cooling is the same?

So E=1/2 μ A2ω2λ
ΔE=E1-E2=1/2 2l μ A2(2Π)2(f12-f22)=3*10^-9J

Is this correct? It seems very small. :)

13. Mar 11, 2015

haruspex

Is that right? I thought it would be 1/4 μ A2ω2l = (1/8) μ A2ω2λ.
Yes, that does seem much too small. Using your equation, I get 3*10-5, so I suspect a powers-of-ten error. What do you get for E1?
By the way, because the expression involves the difference of two close numbers (f12-f22) there will be a loss of precision. You can either calculate the new frequency to several decimal places or, by keeping everything symbolic until the final step, avoid taking a difference of two large numbers.

14. Mar 12, 2015

gasar8

Uf, yes it is 1/4, I searched the formula on the internet, but overlooked that formula that I used is for one whole wavelength.
So:
μ = m/l = ρ*V/l = S*ρ
A = 2*10^-3 m
l = 0,5m
ω = 2Πf
f=√(F1/(ρS))/2l; f2=(F1/(ρS))/4l2
F1-F2=0,3N

ΔE=E1-E2=1/4 μ A2 l (ω12 - ω22)
=1/4 μ A2 l (2Π)2 ((F1/(ρS))/4l2-(F2/(ρS))/4l2)
=1/4 S ρ A2 l 4Π2 * 1/(ρS 4l2) * (F1-F2)
=(AΠ)2/(4 l) (F1-F2)
=(2*10-3m*Π)2/(4*0,5m) * 0,3N
=5,9*10-6 J

15. Mar 12, 2015

haruspex

Yes, that looks better. Haven't checked the numbers in detail, but it's about right.