1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Constrained mass on a spring

  1. Nov 2, 2016 #1
    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. jcsd
  3. Nov 2, 2016 #2
    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
  4. Nov 4, 2016 #3
    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
  5. Nov 4, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  6. Nov 4, 2016 #5
    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.
     
  7. Nov 6, 2016 #6
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Constrained mass on a spring
  1. Mass on a spring (Replies: 4)

  2. Spring and Mass (Replies: 2)

  3. Mass on a spring (Replies: 6)

  4. Mass on a spring (Replies: 14)

Loading...