1. Since , the force determines a potential

   

   Choosing gives

   

   which looks like

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   .

   By inspection the motion is always bound, oscillating between turning points , where . There is a point of stable equilibrium at x=0, where V'(x) = -F(x) =0. At this point,

   

   so we have a restoring force with spring constant , and frequency .

2. 1. For the hanging cable, the force determines a potential

      

      choosing reference point . Over the range of validity , this looks like

      

   2. Using potential methods, the velocity v(x) is given by

      

      For motion which starts at rest at , the energy is all potential at , giving

      

      Thus the velocity is given by

      

      This agrees with the result in Homework 1, 3b.

   3. We attain the greatest final velocity by maximizing the final kinetic energy, by starting with the greatest energy E. This occurs when , E = 0. The final velocity attained by the cable in this case is

      

      Note that this is times the velocity attained by a point mass m falling height l off the table; the point mass converts initial potential energy mgl to final kinetic energy giving . This makes sense as the cable's downward force, doing work to speed up the cable, is (since F = mgx/l) always less than the downward force mg acting on the point particle.

3. To simplify the algebra, let's rewrite V in terms of the dimensionless variable y = x/d. This gives

   

   1. Equilibrium points occur when

      

      Gathering terms in the numerator gives

      

      with equilibrium points at . (The imaginary solutions are unphysical).

      To find stability,we must examine

      

      Using the quotient rule for the derivative, , gives after ample use of the product rule,

      

      This looks horrific, but we really only need its values at the equilibrium points . Note that the final product in the numerator vanishes at both equilibrium points. The first factor in the numerator, containing a sum of 3 terms, is also simple at the equilibrium points, as terms 2 and 3 vanish at y=0, while terms 1 and 2 vanish at . Thus we have values at equilibrium

      

      This means that the origin, with negative (hence negative , by the chain rule), is an unstable equilibrium point, or maximum of V. , with positive , are stable equilibrium points, and minima of V.

   2. To sketch the potential, note that V is even in y, with limiting behavior for . Starting from minus infinity, V(y) decreases to its minimum value at . Plugging into V(y), this value is . V(y) then increases to the local maximum at y=0, where V(y=0) = -W/8. It then decreases to the next minimum at , where V(y=0) = -W/8, then increases to its asymptotic value . Thus the potential looks like

      

      Motion is thus unbounded for E;SPMgt;0. For , the motion is bounded with two turning points . The motion oscillates between and forever. For , the motion is again bounded and we have 4 turning points as shown for the E= -3W/16 case. Here the only allowed motions are bounded oscillations between and , and between and .

   3. Our turning points occur where the energy is all potential, thus

      For the E= -W/16 case, with turning points at :

      

      This gives the quadratic equation for :

      

      with solution, by the quadratic equation,

      

      and an unphysical solution . Thus our turning points are at and the only allowed motion oscillates between x = - 4.06 d and x = + 4.06 d.

      For the E= -3W/16 case, we have 4 turning points, for each x = d y solving

      

      This gives the quadratic equation for y:

      

      with solution, by the quadratic equation,

      

      This has four physical solutions, and . The two allowed motions are both bounded oscillations, one between x = -2.2 d and x = -0.75 d, the other between x = 0.75 d and x = 2.2 d.

   4. A particle can escape to infinity only if it has , which is here the asymptotic value 0. Thus a particle with exactly the minimum necessary escape velocity must have E= 0.

      The particle at the origin has, from part (b), V = -W/8. Since K + V = E = 0, the kinetic energy must be W/8. Thus

      

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