# Homework Help: Fundamental frequency change because of bouyant force

Tags:
1. Jul 15, 2016

### Soren4

1. The problem statement, all variables and given/known data
A rope has an end fixed and the other is passing through a pulley and has a body attached to it. The fondamental frequency of the rope is initially $f_1=400 Hz$. If the body is then put in water the fondamental frequency of the rope becomes $f_2=345 Hz$. If the linear mass density of the rope is $\mu=10 g/m$ determine
a) the density $\rho$ of the body
b) the lenght of the horizontal part of the rope $L$

2. Relevant equations

3. The attempt at a solution
For point a) I don't have problems since
$$\frac{f_2}{f_1}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{mg-\rho_{H_2O}Vg}{mg}}=\sqrt{\frac{mg-\rho_{H_2O}\frac{m}{\rho}g}{mg}}=\sqrt{1-\frac{\rho_{H_2 O}}{\rho}}$$
Where $T$ is the tension of the rope. From which I can get $\rho$

But what about point b)?! It seems impossible to me to get $L$ just knowing $\rho$, $f_1$, $f_2$ and $\mu$..
These conditions are not enough, how can I determine $L$?

2. Jul 15, 2016

### Staff: Mentor

3. Jul 16, 2016

### Soren4

Having the wavelenght would solve the problem because $\lambda_1=\lambda_2=2L$, the problem is that I cannot have $v_1$ and $v_2$ but only the ratio $$\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}$$ Which I already used to determine $\rho$.

Is there another condition I can use to determine $\lambda$?

4. Jul 16, 2016

### haruspex

There is a piece of information you have not used. There is an equation at the link NascentO posted you can use it in.

5. Jul 16, 2016

### Soren4

Thanks a lot for the reply!

I read all the link and formulas but I fail to see the piece of information I did not use. After all here I have 3 indipendent unknown
• Lenght of (horizontal part of) the rope $L$
• Density of the object $\rho$
• Mass of the object $m$ or, equivalently its volume $V$
The thing is that the tension $T$ depends on both $m$ (for weight) and $V$ (for boyant force), or equivalently on $m$ and $\rho$ or also on $\rho$ and $V$. But I need two indipendent pieces of information only for tension.

On the other hand the conditions I have are only 2, that is

$$f_1=\frac{\sqrt{T_1}}{\sqrt{\mu} 2L}=\frac{\sqrt{mg}}{\sqrt{\mu} 2L}\tag{1}$$ $$f_2=\frac{\sqrt{T_2}}{\sqrt{\mu} 2L}=\frac{\sqrt{mg-\rho_{H_20}V g}}{\sqrt{\mu} 2L}\tag{2}$$

And considering 3 indipendent unknowns I really do not know where to get $L$ from.

In the link I found the following formula
$$f=\frac{v}{2L}=\frac{\sqrt{\frac{T}{m_{rope}/L}}}{2L}$$

But here $m_{rope}$ is the mass of the rope which I do not know from data..

Would you be so kind as to suggest me where this not-used piece of information is?

6. Jul 16, 2016

### haruspex

7. Jul 16, 2016

### Soren4

Thanks again for the answer! Knowing $\mu$ allows me to get $L$ once I know $m_{rope}$.

But getting $m_{rope}$ is still a problem, as $m_{rope}$ has nothing to do with the mass of the object $m$, its volume $V$ or its density $\rho$.. So it is still an indipendent variable..

I can surely rewrite the equations I have as

$$f_1=\frac{\sqrt{T_1}}{\sqrt{m_{rope}} 2\sqrt{L}}=\frac{\sqrt{mg}}{\sqrt{m_{rope}} 2\sqrt{L}}\tag{1}$$ $$f_2=\frac{\sqrt{T_2}}{\sqrt{m_{rope}} 2\sqrt{L}}=\frac{\sqrt{mg-\rho_{H_20}V g}}{\sqrt{m_{rope}} 2\sqrt{L}}\tag{2}$$

But these are still 2 equations in 3 variables ($m_{rope}$, $m$ and $V$).

I really do not know how I cannot see this , but it seems to me that the variables are still three and there are just two conditions.. Forgive my incapacity to understand this, if I may ask is there any further clue of the condition to use that you could give me ?

8. Jul 16, 2016

### haruspex

You are not looking at the right equation. In the link NascentO posted, there is an equation relating velocity, tension and mass per unit length.

9. Jul 16, 2016

### Soren4

Thanks a lot for your kind help!

The only possible equation in the link relating these variables is $$v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{T}{m_{rope}/L}}$$ I rewrite the two conditions I have

$$f_1 \cdot 2 L=v_1=\sqrt{\frac{T_1}{\mu}}=\sqrt{\frac{T_1}{m_{rope}/L}}\tag{1}$$
$$f_2 \cdot 2L=v_2=\sqrt{\frac{T_2}{\mu}}=\sqrt{\frac{T_2}{m_{rope}/L}}\tag{2}$$

Again 2 equations and 3 indipendent unknowns ($T_1$, $T_2$, $L$)

This allows me only to find the following ratios (where $L$, or $m_{rope}$ disappear, as they are equal for both cases)

$$\frac{v_1}{v_2}=\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}$$

I tried to use $\mu$ as suggested but it does not give a further indipendent condition

I'm sorry to ask again, but apparently this simple exercise is really driving me crazy in finding this hidden third condition... What use of eq $v=\sqrt{\frac{T}{\mu}}$ and of $\mu$ should I do?

10. Jul 16, 2016

### TSny

Soren4. I'm with you, I don't think there's enough info to find L.

11. Jul 16, 2016

### haruspex

Yes, I'm sorry - I just assumed that because you had not used the value of μ it must be possible to deduce more. But you are quite right, there's not enough information.
Sorry ro have wasted your effort on this.

12. Jul 17, 2016

### Staff: Mentor

Missing data? I hadn't noticed that. Do you have the textbook's answer for L? If so, we could work backwards and determine what's missing....perhaps they intended that you be told the volume of the body to be submerged, e.g., 10 litres.

There's a tiny video clip here that illustrates the change in wave velocity. He uses the rung of a ladder as the pulley from which to hang weights, and then counts each reflection as the wave returns.