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PGRE question: angular freq of small oscillations

  1. Aug 11, 2014 #1
    Problem from past PGRE:

    A particle of mass m moves in a one-dimensional potential V(x)=-ax2 + bx4, where a and b are positive constants. The angular frequency of small oscillations about the minima of the potential is equal to:

    Answer is 2(a/m)1/2.

    I understand how this is found 'the long way' by actually applying calculus, but if you plug in a=-1/2 k and b=0 shouldn't you get back sqrt(k/m) ? Because the harmonic potential is 1/2kx^2 ... so where is the flaw in my logic?

    Last edited: Aug 11, 2014
  2. jcsd
  3. Aug 11, 2014 #2


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    Homework Helper

    If the potential was 1/2 kx2+bx4, there would be oscillation about x=0. But the problem says that both a and b are positive constants. So a can not be equal to -1/2 k where k is positive.

    In case a>0 and b>0, the potential function has maximum at x=0. You do not get oscillatory motion about x=0. The minima are at x=±√(a/2b). Oscillatory motion is possible about a minimum of the potential function.
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