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Verbatim question for GR8677 #60
Mechanics}Simple Harmonic Motion


Recall F=-\frac{\partial V}{\partial x}=-2bx. Simple harmonic motion has the simple form, \ddot{x}\propto x. Thus F=m\ddot{x}=-2bx \Rightarrow \ddot{x}=-\omega^2 x, where \omega^2=2b/m is the frequency squared. Thus, simple harmonic motion occurs with a frequency determined by both b and m. This is choice (C).

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Comments
ernest21
2019-08-23 02:03:37
I enjoyed reading your Solutions together with the answers of others. This is truly a great read for me. brick breaker 2NEC
a62
2016-09-20 02:36:59
Easy, but even easier if you look at it this way: a is absolutely meaningless because the zero of potential is arbitrary. So it\'s exactly like the mass-on-a-spring model.NEC
Almno10
2010-11-11 23:02:29
Force is the negative gradient of potential energy. The a drops out.
psychonautQQ
2013-09-26 08:41:23
but this is asking for the frequency, how does force and frequency relate?
NEC
sirius
2008-11-05 22:26:45
A similar solution if you remember V(x) = \frac{1}{2}m\omega^2x^2. The given V(x) has a vertical shift, a, which can be ignored by shifting your zero-point energy. So, b=\frac{1}{2}m\omega^2, solving for \omega makes it depend on b and m.

This is enough, but the problem asks for frequency f=\frac{\omega}{2\pi}. So f=\frac{1}{2\pi}\sqrt{\frac{b}{2m}}. making f depend on b and m. The answer then is (C).
alemsalem
2010-09-21 06:25:37
i totally agree, to be sure that it doesn't depend on a, just remember that shifting the potential energy by a constant cannot change the motion (classically) so it doesn't affect the frequency.
NEC

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A similar solution if you remember V(x) = \frac{1}{2}m\omega^2x^2. The given V(x) has a vertical shift, a, which can be ignored by shifting your zero-point energy. So, b=\frac{1}{2}m\omega^2, solving for \omega makes it depend on b and m. This is enough, but the problem asks for frequency f=\frac{\omega}{2\pi}. So f=\frac{1}{2\pi}\sqrt{\frac{b}{2m}}. making f depend on b and m. The answer then is (C).

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