Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Energy & initial conditions

  1. Jul 2, 2010 #1
    The total energy of a particle is given by:

    [tex]E_{tot} = 2 \dot{x}^{2} - cos(\frac{1}{2} \pi x)[/tex]

    and i'm told that the particle passed through the point [tex]x=1 [m][/tex] with a velocity of [tex]\vec{v}=-\frac{1}{2} \hat{x} [m/sec][/tex].

    I'm required to find the maximum velocity of the particle during its motion.

    The attempt at a solution
    well I know that at equilibrium, the total energy of the particle is kinetic energy, so this corresponds to the maximum velocity.
    Equilibrium occurs when the force on the body, i.e. the first derivative of the potential energy term with respect to x, vanishes.
    The P.E. is extracted from the total energy expression, that is,

    [tex] U(x)= - cos(\frac{1}{2} \pi x) [/tex]


    U'(x) = 0 ===> equilibrium points are: [tex]x_{e} = 2n , n=0,1,2,... [/tex]

    and at this point i'm stuck!
    Is it even the right approach to solve questions like these, or not?!

    by the way, the answer is:

    [tex] v_{max} = \sqrt{\frac{3}{4}} [m/sec] [/tex]
  2. jcsd
  3. Jul 2, 2010 #2
    That approach looks correct. What you gotta do now is to figure out Etot which can be determined from the givens (just plug in x, and x'). Then with the equilibrium x, Etot, solve for x' to get the answer.

    actually you should check the values of x for U' = 0 because they may be maximas as well e.g. if x is 2, which doesn't even given a solution.
    Last edited: Jul 2, 2010
  4. Jul 3, 2010 #3
    I assume that the total energy is conserved, which is an important assumption in this problem.
    First, I think you cannot directly deduce U(x) from the total energy (yes, just by looking at the form of the formula, we can see it; but strictly speaking, we cannot conclude without proof). So here is the way: because the total energy is conserved, dE/dt=0. From this, you can deduce the force, and then, U(x) (remember that the potential energy only goes with a conservative force).

    From here, you have 2 ways:
    1 - The same way you approach the problem. Remember that E is conserved, so plug the data they give in to find E.
    By the way, I don't think the value x=2 is not appropriate due to that it's a maximum. It is because the particle doesn't have enough energy to reach x=2.
    2 - Use x"(t) (from the force) to deduce v(x).

    Just my 2 cents :smile:
  5. Jul 3, 2010 #4
    That's true, though even if Etot is large enough such that a solution exists for x=2, that's still not the right answer because U(x) is not at the minimum for x=2.

    A easy way to get U'(x) might be to consider Etot = K(x') + U(x), then get delE/delx = U'(x) = 0
  6. Jul 3, 2010 #5
    Thanks guys!

    A question:

    There is no need to "think of" conservative forces, isn't it?! as if I follow what has been just said, then:

    [tex] \frac{dE_{tot}}{dx} = \frac{dU}{dx} = \frac{1}{2} \pi sin (\frac{1}{2} \pi x )[/tex]

    so I only need to integrate this expression to obtain the P.E. without even recognizing that this is the force, had it been conservative ; that is,

    [tex]U(x) = cos(\frac{1}{2} \pi x_{0}) - cos(\frac{1}{2} \pi x)[/tex]

    and plugging in the Initial place of the particle [tex]x_{0} = 1 [m][/tex] yields the expected expression for the P.E.

    [tex]U(x) = -cos(\frac{1}{2} \pi x ) [/tex]

    while the stable equilibrium points correspond to [tex] \frac{d^{2}U}{dx^{2}}>0 [/tex] ... and afterwards I continue like you guys have just said to obtain the max. velocity.
  7. Jul 3, 2010 #6
    The potential energy only exists when there is conservative force, that's what you have to point out, though it's pretty obvious in this problem.
  8. Jul 3, 2010 #7
    Also I don't think you need to integrate U' to obtain U, because U' and U" is really all you need to find the x for which U is at minimum (KE at maximum).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook