- #1
MalachiK
- 137
- 4
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?
Hooke's law etc.
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.
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?
Homework Equations
Hooke's law etc.
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.