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Homework Help: 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


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    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?
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