Tension in string and oscillation

[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

Homework 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)

 

[Orodruin]

Do you know how wave velocity depends on string tension?

 

[gasar8]

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

 

[haruspex]

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

Homework 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)

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).
Please post the details.

 

[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 α_s = 1.2*10^-5 / K for the string and α_f = 1.7*10^-5 / K for the frame.

So:
Δl_s/l = F/(ES)+α_s ΔT
Δl_f/l = α_f ΔT

Δl_s = Δl_f
F/(ES)+α_s ΔT= α_f ΔT
F=(α_f - α_s) ΔT E S
F= -0,3N --> What does this mean?

 

[haruspex]

F/(ES)+α_s ΔT= α_f ΔT
F=(α_f - α_s) ΔT E S
F= -0,3N --> What does this mean?

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

 

[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? :)

 

[haruspex]

F1=120,8N

That's double what it should be. Did you take l as 1 instead of .5?

This result is an octave higher which seems a little bit exaggerated

It certainly shouldn't be much changed.

Do I even have to take into account that l2=l-(αf ΔT-αs ΔT), because it is negligibly small?

The change in length is not (αf ΔT-αs ΔT). But you are right, it does not have much effect.

 

[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?

 

[haruspex]

I get final frequency f2=437 Hz?

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

 

[gasar8]

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

 

[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. :)

 

[haruspex]

E=1/2 μ A²ω²λ

Is that right? I thought it would be 1/4 μ A²ω²l = (1/8) μ A²ω²λ.

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

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.

 

[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

 

[haruspex]

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

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