Can someone tell me how to approach this problem?

Find the period of oscillation of a particle of mass m, constrained to move along the x-axis, with a potential V(x)= a+bx+cx2 for a small disturbance from the equilibrium position.
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 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Hi Idoubt! What equations do you know relating the potential to the motion?

 Quote by tiny-tim Hi Idoubt! What equations do you know relating the potential to the motion?
well only thing I can think of here is that F=-2cx-b
and also that T+U=E ( T= K.E, U=P.E, E=total energy) but this isn't in the simple harmonic oscillator form F=-kx so I don't know how to proceed from here.

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 Quote by tiny-tim How about a differential equation?
Well I thought about expressing the force equation in the terms of a phase angle like

x=x0cosa where x0 is the amplitude, and integrating and solving for a = 2pi , but uh I got a soup that I couldn't integrate
 I solved the basic force equation md2x/dt2=-2cx-b and got the solution x=Acos((2c/m)1/2t)+Bsin((2c/m)1/2t)+2c/b In this it seems to me that (2c/m)1/2 is w the angular velocity and if so since w=2pi/T the time period should pop out as T=2pi(m/2c)1/2 , does that seem right?
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor yup … that's exactly right! (btw, you could rewrite md2x/dt2=-2cx-b as d2(x+b/2c)/dt2=-(2c/m)(x+b/2c), and in that form you can just read off T without having to find the general solution! )

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 Quote by Idoubt Can someone tell me how to approach this problem? Find the period of oscillation of a particle of mass m, constrained to move along the x-axis, with a potential V(x)= a+bx+cx2 for a small disturbance from the equilibrium position.
 Quote by Idoubt well only thing I can think of here is that F=-2cx-b and also that T+U=E ( T= K.E, U=P.E, E=total energy) but this isn't in the simple harmonic oscillator form F=-kx so I don't know how to proceed from here.

V(x)= a+bx+cx2 is a parabola with a minimum (assuming that c>0) at x=-b/(2c). x=-b/(2c) is the equilibrium position.

Factror F=-2cx-b to get:

$$\displaystyle F=-2c\left(x-\left(-{{b}\over{2c}}\right)\right)$$

This is in simple harmonic oscillator form. 2c is analogous to the spring constant, k.

You have, F = -k*u, where u=x+(b/(2c)) and k=2c.
 Thank you both, that makes it a lot clearer. Here the term x+b/2c is the actual displacement from the equilibrium position right? so the formula F=-kx is applicable only what x=0 is the equilibrium position.
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor yes! and yes!
 Recognitions: Science Advisor I don't know if it will help you or confuse you more, but Lagrangian formalism is very useful for dealing with problems of motion in given potential. Take a look at Wikipedia article. If it looks too scary, never mind.