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Two Mechanics Questions 83, 84

  1. Nov 30, 2003 #1
    There is a diagram that accompanies it so if the explanation isn't clear you can refer to the sample test posted on

    the GRE.org website.

    83 on GRE. Consider a particle moving without friction on a rippled surface, as shown above. Gravity acts down in the negative h direction. The elevation h(x) of the

    surface is given by h(x) = d cos[kx]. If the particle starts at x=0 with a speed v in the x direction, for what values of v will the particle stay on the

    surface at all times?

    The answer is v <= Sqrt[g/(k^2*d)]
    Why, and what concepts are involved here?

    84. Two pendulums are attached to a massless spring, as shown above. The arms of the pendulums are of identical lengths l, but the pendulum balls have unequal

    masses m1 and m2. The initial distance between the masses is the equilibrium length of the spring, which has spring constant K. What is the highest normal

    mode frequency of the system?
    A. Sqrt[g/l]
    B. Sqrt[K/(m1+m2)]
    C. Sqrt[K/m1 + K/m2]
    D. Sqrt[g/l + K/m1 + K/m2]
    E. Sqrt[2g/l + K/(m1+m2)]

    The answer is:
    D. Sqrt[g/l + K/m1 + K/m2]
    My question is why, and how do you know that this frequency is the highest?
  2. jcsd
  3. Nov 30, 2003 #2
    If the speed is too high, the particle will lift off because of the centrifugal force. You can calculate the speed by the conservation of energy. You can find the centrifugal force if you know the radius of curvature of the path. Remember the lift-off can only happen where the path is convex (curved downward).
    I think it's obvious that the motion depends on both the values of g and K, because gravity acts against/with the spring force. So only D or E can be correct. Now imagine K was zero, then surely you have two independent pendulums with period sqrt(g/l) each. Thus, D must be correct.
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