Oscillation in quadratic potential

[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.

 

[tiny-tim]

Hi Idoubt!

What equations do you know relating the potential to the motion?

 

[Idoubt]

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.

 

[tiny-tim]

How about a differential equation?

 

[Idoubt]

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

 

[Idoubt]

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?

 

[tiny-tim]

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! )

 

[SammyS]

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.

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.

 

[Idoubt]

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.

 

[tiny-tim]

yes! and yes!

 

[K^2]

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.