# The sound of two wires

1. Jan 4, 2014

### Numeriprimi

The problem statement, all variables and given/known data
The rod of length l and mass m_0 is hinged on two same wires (Fig. 3 on second page - example ,,4. Zvučící dráty" - http://fyzikalniolympiada.cz/archiv/55/fo55a1_z.pdf). With this load, each of the wires gives basic tone height c in natural tuning, f_c = 264 Hz.
A) Move the left wire to the right wire (Fig.4). Now wires sounding in fourth interval: f(left wire)/f(right wire) = 4/3. Determine the length x_1 of move and the frequencies f_1 and f_2.
B) Back to the initial situation - now hing a weight on rod (Fig. 4). Than left wire sounding in tone e a nd right wire sounding in tone g. YOU KNOW: f_e/f_c = 5/4 and f_g/f_c = 3/2. The frequency of tone is proportional to the square root of the tension force.
Determine the length x_2 and the m of weight.

The attempt at a solution
A) Determine the frequencies f_1(left) and f_2(right): 4/3 = f_1/f_2; f_2 is still on c, so f_2 = f_c = 264 Hz; f_1 = 4/3 * f_2 = 4/3 * 264 Hz = 352 Hz
Determine the lenght x_1 ... I have a strange solution, may wrong.
f_1(left)/f_2(right) = distance of right wire to the left end of the rod / distance of left wire to the right end of rod = l/(l-x_1)=4/3; 3l = 4l - 4x_1; l = 4x_1; x_1 = l/4, so, this is my resoult.

B) We what know what is f_e/f_g... f_e/f_g = f_e/f_c : f_g/f_c = f_e/f_g = 5/4 : 3/2 = 5/6
The frequency of tone is proportional to the square root of the tension force... f= k*√F
f_e/f_g = (k*√F_1)/(k*√F_2) = √(F_1/F_2) = 5/6
F_1/F_2 = 5/6 * 5/6 = 25/36 ... So, this is ratio of the acting forces. I don't know more. Can you help me and check my calculations? Thank you very much and sorry for my bad English :-)

2. Jan 4, 2014

### voko

What is the equation for the fundamental frequency?

3. Jan 4, 2014

### Numeriprimi

Which frequency? Some different formula? Ok, i can express frequency f=1/T=v/λ.
And fundamental frequency is the lowest frequency, other frequencies are natural multiples. I don't know more.

4. Jan 4, 2014

### voko

There is a formula that relates a string's fundamental frequency to its tension and other things. You will need it for this problem.

5. Jan 4, 2014

### Numeriprimi

Hmmm... I know some formulas for oscillation.
First is from Newton's law: F= ma = - ωy^2 * m, so I can write F = - 2πf * y^2 * m
Second is F= - ky, I think it isn't important for this example.
So, the first formula is good?
There is f, hmm, ok, but what with it? Ratio? F_1/F_2 = (-2πf_1* y_1^2 * m)/(-2πf_2* y_2^2 * m)
F_1/F_2 = (f_1*y_1^2)/(f_2*y_2^2) ... and F_1/F_2 = 25/36 from my previous calculations.
Hmmm, I can have some system of equations... But what can I do with y?

6. Jan 4, 2014

### Numeriprimi

And the part A is all good or no?

7. Jan 4, 2014

### voko

I have re-read you formulation again. You did write: "the frequency of tone is proportional to the square root of the tension force". That is enough for this problem. What you need is to determine tension in the wires in both cases.

Note that the wires and the rod are in static equilibrium.

8. Jan 4, 2014

### Numeriprimi

But how? I really don't know how, if I know only ratio between two forces.

9. Jan 4, 2014

### voko

The ratio should be sufficient. You know the basic tone. It corresponds to the initial tension. After the rod/wire configuration is re-arranged, the forces of tension change. You can find the ration with each other and also with the original force.

10. Jan 4, 2014

### Numeriprimi

F_1*d_1 = F_2*d_2; F_1*x_2 = F_2*(l-x_2); F_1/F_2 = (l-x_2)/x_2 = l/x_2 -1
(F_1/F_2 +1)*x_2 = l
x_2 = l/(F_1/F_2+1)= l/(25/36+1) = l/1,7 ... it is right?

11. Jan 4, 2014

### voko

I do not understand what your symbols mean. I understand $l$ (length of the rod) and $x_2$ (position of the additional mass). The rest are mystery. Please explain what they are, and also explain how you obtain the equations.

12. Jan 4, 2014

### Numeriprimi

d_1 ... distance from the left end of the rod to weight; F_1 ... force acting on the left end of the rod
D_2 ... distance from the right end of the rod to weight; F_2 ... force acting on the right end of the rod
F_1 * d_1 = F_2 * d_2 (generally eg levers, it can be likened?)
d_1 = x_2
d_2 = l - x_2
x_2 * F_1 = F_2 * (l-x_2); F_1/F_2 = (l-x_2) / x_2 = l/x_2 - 1
(F_1/F_2 +1) * x_2 = l
x_2 = l / (F_1/F_2 +1) = l / (25/36 +1) = l / 1,7

13. Jan 4, 2014

### Numeriprimi

So, it is OK or no? :-)

14. Jan 5, 2014

### voko

I do not think you are approaching it correctly. You should be considering the equations of equilibrium to find the forces of tension, but you are not doing it. For example, in the basic case, there are three forces: $T_{0l}$, tension in the left wire, $T_{0r}$, tension in the right wire, and $P = -mg$, the weight of the rod. In equilibrium, the sum of all forces must be zero: $T_{0l} + T_{0r} + P = 0$. The sum of the moments of forces must also be zero, so taking the moments about the center of the rod, we have: $- T_{0l} l/2 + T_{0r} l/2 = 0$. Thus we have $T_{0l} = T_{0r} = T_0$. We further have $2 T_0 = - P$, so $T_0 = mg / 2$. We are given that $f_0 \propto T_0^2$, so $$f_0 \propto {(mg)^2 \over 4}.$$ You can deal with the other two cases in a similar manner.

15. Jan 6, 2014

### Numeriprimi

Thank you very much, I have the resoults, it was a system of equations with three unknowns... Thanks! :-)