# Homework Help: Constrained mass on a spring

1. Nov 2, 2016

### MalachiK

1. The problem statement,
A spring of stiffness q and natural length l0 is fixed at one end to a point x = 0, y = l0 and at the other end to a mass m that is constrained to move horizontally and displaced through a horizontal distance x. The length of the string in this position is l.

1. For x << l0, how does the force in the horizontal direction change with x?
2. If the potential U(x) ≈ Axn, for small x, what are the values of A and n in terms of the constants given?

2. Relevant equations
Hooke's law etc.

3. The attempt at a solution
It seems to me that the force on the string should be q(l - l0) in the direction of the spring and that the horizontal component of this is q(l - l0) cos(θ) where θ is the angle between the spring and the horizontal. To get something explicitly in x I need to unpack this a bit.

l = √(l02 + x2) and cos(θ) = x/l I get that...
$F_x = \frac{q(\sqrt{l_0^2+x^2}-l_0)x}{\sqrt{l_0^2+x^2}}$
and after squaring, cancelling off terms and taking the root again becomes
$F_x = \frac{qx^2}{\sqrt{l_0^2+x^2}}$

I think that if x << l0 we can say that $l_0^2 + x^2 \approx l_0^2$
So $F_x = \frac{qx^2}{l_0}$

q/l0 has units of N/m2.. giving my force expression units of Newtons so my heart fills with hope that it might be correct! It's too good to be true, so almost certainly wrong.

Now since $\frac{dU}{dx} = F = \frac{qx^2}{l_0}$ I get $U(x) = \frac{qx^3}{3l_0}$ which looks something like what we should have and gives us A in terms of some stuff in the problem statement. BUT.. I don't see how n can be expressed in terms of constants from the original problem. The only two candidates are the length terms that would divide to give a dimensionless exponent. I can't make this connection.

Also, later in the problem we're given n = 4 and asked to find the period of the motion. This makes me suspect that n should in fact depend on something to do with the values in the system but I can't see how.

If everything that I've done is correct I shall keep on thinking (tomorrow - it's past midnight here). But if somebody points out a problem then I'll go back and fix it.

2. Nov 2, 2016

### Hamal_Arietis

OK, but how you make this:
My formula is:
$$F_x = \dfrac{q(\sqrt{l_0^2+x^2}-l_0)x}{\sqrt{l_0^2+x^2}}=\frac{qx^3}{(\sqrt{l_0^2+x^2}+l_0)\sqrt{l_0^2+x^2}}$$
I think you shouldn't use this approximate
I think you should use Taylor series approximate:$$(1+(\frac{x}{l_0})^2)^n$$

Last edited: Nov 2, 2016
3. Nov 4, 2016

### MalachiK

I'm taking another look at this this evening.
$F_x = \frac{q(\sqrt{l_0^2+x^2}-l_0)x}{\sqrt{l_0^2+x^2}}$ is just...
$F_x = q x \frac{\sqrt{l_0^2+x^2}-l_0}{\sqrt{l_0^2+x^2}}$
$F_x = qx \bigg({1-\frac{l_0}{\sqrt{l_0^2+x^2}}}\bigg)$
I don't know what I was thinking the first time I tried to wrangle that expression. I think I got fixated on multiplying to remove the roots or something.
For very small values of x, I guess I can take the first couple of terms from the Taylor series of the fraction
$\frac{l_0}{\sqrt{l_0^2+x^2}} = 1 - \frac{l_0x^2}{2l_0^3} + ...$
and put that into the force expression to get
$F_x = -qx\bigg(\frac{x^2}{2l_0^2}\bigg) = \frac{-qx^3}{2l_0^2}$
This seems like a better answer because it has the force pointing in the right direction, the correct units and it has the potential energy going like $x^4$.

Last edited: Nov 4, 2016
4. Nov 4, 2016

### haruspex

Looks right. Just before the final step, you lost an exponent of 2 on l0 in typing it in, but it reappeared at the end.

5. Nov 4, 2016

### MalachiK

Oh yeah. thanks for pointing that out. I just did an edit and fixed it.

Now I'm going to have a crack at the time period.

6. Nov 6, 2016

### MalachiK

Okay.. setting $B = \frac{q}{2ml_0^2}$ I can write $F_x(x) = m\frac{d^2x}{dt^2}$ so $\frac{d^2x}{dt^2} = -Bx^3$

I had to look up some examples, but in the end I got a solution that I posted here. I should have read the rules and not posted this in the DE forum. Anyhow, eventually I get this...

$\sqrt{\frac{2}{B}}\int{\frac{1}{\sqrt{x_0^4+x^4}}dx} = \int{dt}$

It seems that the integral on the left is only defined for $x < x_0$ which makes we wonder if this is correct. I don't actually need to evaluate the integral as in this case the problem just asks me to show that the period = some expression that looks like the left hand side integral evaluated between some limits. In fact, the result I'm given is [some constants] $\int_0^1{\frac{du}{\sqrt{1-u^4}}}$

If I evaluate the integral on the right between 0 and T/4 then I'll get T/4 (T = the period). This means that the left hand side integral needs to be evaluated between 0 and $x_0$, right?