GR 8677927796770177 | # Login | Register

GR8677 #60
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Mechanics$\Rightarrow$}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).

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
Comments
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.
Almno10
2010-11-11 23:02:29
Force is the negative gradient of potential energy. The a drops out.
 psychonautQQ2013-09-26 08:41:23 but this is asking for the frequency, how does force and frequency relate?
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).
 alemsalem2010-09-21 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.

Post A Comment!
 Username: Password:
Click here to register.
This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

## Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...