# Mass attached on two springs between the walls

1. Jun 16, 2014

### skrat

1. The problem statement, all variables and given/known data
We have two springs each $0.5 m$ long and both have $k=5N/cm$. One end of the spring is attached to the wall while the other two ends are connected in the middle. On the junction point of the two springs a mass is hanged. How does the frequency of oscillation around equilibrium state depend on the initial offset due to the mass?

Hint: There is no need to explicitly write the initial offset as function of mass. Just use the offset as independent parameter using which mass can be calculated.

2. Relevant equations

3. The attempt at a solution

$mg+2Fsin\varphi=ma$

$mg+2k\frac{z}{\sqrt{l^2+z^2}}(\sqrt{l^2+z^2}-l)=m\ddot{z}$

Let's say that $z=D+\varepsilon$ where $\varepsilon$ is very small.

$mg+2k\frac{D+\varepsilon }{\sqrt{l^2+(D+\varepsilon )^2}}(\sqrt{l^2+(D+\varepsilon )^2}-l)=m\ddot{\varepsilon }$

$mg+2k(D+\varepsilon)-\frac{2kl(D+\varepsilon )}{l\sqrt{1+(\frac{D+\varepsilon }{l})^2}}=m\ddot{\varepsilon }$

$mg+2k(D+\varepsilon )-2k(D+\varepsilon)[1-\frac 1 2 (\frac{D+\varepsilon }{l})^2]=m\ddot{\varepsilon }$

$mg+\frac{2k}{l^2}(D+\varepsilon )^3=m\ddot{\varepsilon }$

$mg+\frac{2kD^3}{l^2}[1-3\frac \varepsilon D]=m\ddot{\varepsilon }$

$\ddot{\varepsilon }+\frac{6kD^2}{ml^2}\varepsilon =g+\frac{2kD^3}{l^2}$

So if I am not mistaken, than $\omega \propto D$. Right?

2. Jun 16, 2014

### BvU

You sure you rendered the problem correctly ? Vertical walls ? Vertical oscillation ?
Is there any reason at all you may assume that D << l for your Taylor expansion ?
What if the walls are indeed vertical and 0.8 m apart with springs length 0.5 m each ?

3. Jun 16, 2014

### skrat

Vertical walls? YES.

Vertical oscillations? YES. It's says that explicitly in the problem, I just forgot to translate that part.

Good point. There is a question $b$ to this problem which I did not write because I know how to answer it. And in the question b it says that the body has $50$ grams. So to answer your question: No there is no reason to assume that $D<<l$ if you read only what is written in my original post. However if you also read part b (which you couldn't), than you can find out that the spring has quite large $k$ for so little mass and therefore I think it was safe to assume that $D<<l$.

My inability to speak English properly is the answer to the last question. The problem also states that the springs are NOT deformed before the mass is applied.

4. Jun 16, 2014

### Staff: Mentor

Hi Skrat.

If z represents the downward displacement, then the starting equation should be (note the - sign in front of the spring term):
$$mg-2k\frac{D+\varepsilon }{\sqrt{l^2+(D+\varepsilon )^2}}(\sqrt{l^2+(D+\varepsilon )^2}-l)=m\ddot{\varepsilon }$$.
Now for the approximation:
$$\sqrt{l^2+(D+\varepsilon )^2}-l≈\frac{(D+ε)^2}{2l}$$
So
$$mg-k\frac{(D+ε)^3}{l^2}=m\ddot{\varepsilon }$$
So my term for the magnitude of the spring force is half of yours.
If I continue through the analysis, I get:

$$mg-k\frac{D^3+3D^2ε}{l^2}=m\ddot{\varepsilon }$$

Notice the + sign, compared to the - sign in your equation. Note that ε = 0 defines the equilibrium position. So,

$$mg-k\frac{D^3}{l^2}=0$$
and
$$-3k\frac{D^2ε}{l^2}=m\ddot{\varepsilon }$$
What happens if you solve the first equation for D in terms of l, and substitute this into the second equation?

Chet

5. Jun 17, 2014

### skrat

I have absolutely no idea how you found this out. OR even less why my method didn't bring me to the same result with a sign error.

Ok, I agree with $$\sqrt{l^2+(D+\varepsilon )^2}-l≈\frac{(D+ε)^2}{2l}$$ which than leaves you with

$mg-\frac{2k(D+\varepsilon )}{\sqrt{l^2+(D+\varepsilon )^2}} \frac{(D+\varepsilon)^2}{2l}=m\ddot{z}$

Now I am assuming you tried to approximate $\frac{1}{\sqrt{l^2+(D+\varepsilon )^2}}$. I don't know how you did it but my way brings me to a completely different conclusion:

$\frac{1}{\sqrt{l^2+(D+\varepsilon )^2}}=\frac{1}{l\sqrt{1+(\frac{D+\varepsilon }{l})^2}}≈\frac 1 l (1-\frac{1}{2} (\frac{D+\varepsilon }{l})^2+...)$

Meaning:

$mg -2k(D+\varepsilon )\frac 1 l (1-\frac{1}{2} (\frac{D+\varepsilon }{l})^2)\frac{(D+\varepsilon)^2}{2l}=m\ddot{z}$

$mg-\frac{k(D+\varepsilon )^3}{l^2}+\frac{k(D+\varepsilon )^5}{4l^4}=m\ddot{z}$

Now It looks like you completely ignored the last term. Why would you do that? It also includes $ε$ just as the second term does.

Now to answer your last question $-3k\frac{D^2ε}{l^2}=m\ddot{\varepsilon }$

$m\ddot{\varepsilon }+3k\frac{D^2ε}{l^2}=0$

$\ddot{\varepsilon }+3k\frac{1}{ml^2}(\frac{mgl^2}{k})^{2/3}ε=0$

right?

6. Jun 17, 2014

### skrat

:)

Btw, thanks for all the help so far

7. Jun 17, 2014

### Staff: Mentor

Within the framework of the approximation we've made, that term is negligible compared to the second term because $\frac{1}{2} (\frac{D+\varepsilon }{l})^2<<1$

I haven't checked your algebra, but I'm assuming it's correct. But, I think you should re-do the coefficient of ε so that the exponents on each of the parameters are combined to a single exponent.

Chet

8. Jun 19, 2014

### BruceW

I've been looking at this thread, but I still have no idea what the set-up is meant to be... There are two vertical walls either side of a mass, which is held there because of two springs (each of which is attached to the mass and to one of the walls). And the idea is that the mass undergoes vertical oscillations. Did I understand it all OK? If so, I don't understand the reason for your starting equation...

9. Jun 19, 2014

### BruceW

I've attached a picture of what I think the experimental set-up is. Have I understood it correctly? p.s. the two wiggly lines are meant to be springs.

#### Attached Files:

• ###### springs.png
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10. Jun 20, 2014

### skrat

You have.

So, what exactly is supposed to be wrong with my starting equation?

11. Jun 20, 2014

### BruceW

Generally, the force on the mass due to a spring is: $\vec{F}=-k\vec{d}$ where $\vec{d}$ is the displacement vector pointing from the equilibrium position of the spring, to the mass itself. Or in this case, you can just think of $-\vec{d}$ as having a direction which points from the mass, towards the place where the spring attaches to the wall. So now if you have two springs attached to opposite walls, what do you get by adding the two forces together? (If you think about it this way, it really seems like your starting equation does not make sense. So I think one of us has not got the right starting equation).

12. Jun 20, 2014

### skrat

Twice as much force on the same mass? But I have factor 2 in my starting equation. I guess you're trying to warn me about the sign error in my post #1. If that's the case, Chet already mentioned that in post #4.

Or maybe I completely misunderstood what you are trying to say?

13. Jun 20, 2014

### ehild

ehild

14. Jun 20, 2014

### BruceW

hmm. OK, let's say $z$ is in the 'down' direction (similar to how skrat has defined it). And say $l$ is half the distance between the two walls. So the force due to the left spring is: $-k(z\hat{z}+l\hat{x})$ (where $\hat{x}$ is the horizontal direction). And the force due to the right spring is: $-k(z\hat{z}-l\hat{x})$. And the force due to gravity is simply $mg\hat{z}$ with $g$ being 9.8m/s^2. So the resultant force is:
$$\vec{F} = -k(z\hat{z}+l\hat{x}) -k(z\hat{z}-l\hat{x}) + mg\hat{z}$$
And now, cancelling some terms, and assuming that the only acceleration is in the z direction, we get:
$$m\ddot{z} = -2kz + mg$$

edit: no wait, $z\hat{z}+l\hat{x}$ is the displacement from where the left spring attaches to the wall. So the displacement from the equilibrium position of the spring would instead be
$$z\hat{z}+l\hat{x}-l\frac{z\hat{z}+l\hat{x}}{\sqrt{z^2+l^2}}$$
And adding this to the force due to the other spring gives:
$$-2kz\hat{z}(1-\frac{l}{\sqrt{z^2+l^2}})$$
OK, I see. yes, your starting equation is the correct one. ah, sorry about that. I forgot the displacement is from the equilibrium point of the spring, not the wall !

Last edited: Jun 20, 2014
15. Jun 20, 2014

### BruceW

OK, sorry about all that. I have one suggestion that is hopefully helpful. Since they don't explicitly say in the question to make $D<<l$ Then I think you should not make this approximation. Just keep the lower order terms in $\varepsilon$. And then at the end, you can always choose whether to make the approximation $D<<l$ or not. But since they don't explicitly say to use this approximation, my advice would be to not make this approximation.

16. Jun 20, 2014

### BvU

@ehild: he did in post #1! and agreed to the minus sign for the k term in post #5.

All we worry about now is what to do with the $m\ddot{\varepsilon }+3k\frac{D^2ε}{l^2}=0$.

Chet is very quiet, so I'll chip in for him: for D we had (from $\epsilon = 0$, equilibrium): $k\frac{D^3}{l^2}=mg$. Well, his tip in post #7 is to be interpreted a little different from what skrat does in post #5. Chet wants to write $\ddot \epsilon + {3g\over D}\epsilon=0\$ so $\ \omega^2={3g\over D}$ from which the final answer $\omega\propto{1\over\sqrt{D}}\$ or $\ f = {1\over 2\pi} \sqrt{3g\over D}\$.

This is really nice, since for the much simpler case of one vertical spring with a weight the expression is the same, except the 3 is then a 1 !

At this point I wonder what part b of the exercise asks, since it wasn't worth mentioning because skrat already knew how to solve it...

17. Jun 20, 2014

### skrat

I more or less got the same result in #5, but I agree that yours looks way better.

$\ddot{\varepsilon }+3k\frac{1}{ml^2}(\frac{mgl^2}{k})^{2/3}ε=0$
$\ddot{\varepsilon }+3k\frac{1}{ml^2}(D^3)^{2/3}ε=0$
$\ddot{\varepsilon }+3\frac{k}{ml^2}D^2ε=0$

We know that $D^3=\frac{gl^2m}{k}$ so $\frac{k}{ml^2}=\frac{g}{D^3}$

$\ddot{\varepsilon }+3\frac{g}{D^3}D^2ε=0$

$\ddot{\varepsilon }+\frac{3g}{D}ε=0$

Part b is the same as here: https://www.physicsforums.com/showthread.php?p=4777170#post4777170

So part b wants me to write $z(t)$ if we release the mass from the same height where none of the two springs is deformed. It is the same problem as given on the link above.

18. Jun 20, 2014

### skrat

HAHA now I see what bothers you. :D

The problem wants me to find $\omega (D)$ but I completely got rid of $D$ in #5 instead. Silly me. -.-

19. Jun 20, 2014

### BruceW

I think ehild was replying to me, not to skrat.

p.s. I'm still concerned about using the approximation $D<<l$ Because even though they say to use this approximation in part b), if they don't say to use it in part a), then I think you shouldn't use it for part a).

20. Jun 20, 2014

### ehild

Yes, and find $\omega (D)$ by the Taylor expansion about D, as Bruce suggested.

ehild