Oscillation in quadratic potential

In summary: That's great to know! Thanks!In summary, the potential V(x)=-a+bx+cx2 determines the motion of a particle, subject to a small displacement. To find the period of oscillation, one must solve for x=Acos((2c/m)1/2t)+Bsin((2c/m)1/2t)+2c/b. This gives x=Acos((2c/m)1/2t)+Bsin((2c/m)1/2t)+2c/b=T, where T=2pi(m/2c)1/2.
  • #1
Idoubt
172
1
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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  • #2
Hi Idoubt! :smile:

What equations do you know relating the potential to the motion? :wink:
 
  • #3
tiny-tim said:
Hi Idoubt! :smile:

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

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.
 
  • #4
How about a differential equation? :smile:
 
  • #5
tiny-tim said:
How about a differential equation? :smile:

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 :redface:
 
  • #6
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?
 
  • #7
yup … that's exactly right! :biggrin:

(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! :wink:)
 
  • #8
Idoubt said:
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.

Idoubt said:
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:

[tex]\displaystyle F=-2c\left(x-\left(-{{b}\over{2c}}\right)\right)[/tex]

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.
 
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  • #9
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.
 
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  • #10
yes! :wink: and yes! :smile:
 
  • #11
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.
 

1. What is oscillation in a quadratic potential?

Oscillation in a quadratic potential refers to the behavior of a system where the position or value of a variable changes repeatedly between two or more values, within a quadratic potential energy function.

2. How does a quadratic potential affect oscillation?

A quadratic potential can affect oscillation by providing a restoring force that brings the system back to its equilibrium position. This force is proportional to the displacement from the equilibrium and causes the system to oscillate around the equilibrium point.

3. What is the equation for describing oscillation in a quadratic potential?

The equation for oscillation in a quadratic potential is known as the harmonic oscillator equation, given by F = -kx, where F is the restoring force, k is the spring constant, and x is the displacement from equilibrium.

4. What are some real-world examples of oscillation in a quadratic potential?

Oscillation in a quadratic potential can be observed in many physical systems, such as a mass attached to a spring, a pendulum, a guitar string, or a tuning fork. It is also seen in electronic circuits, where the voltage and current exhibit oscillatory behavior.

5. How does the amplitude of oscillation change in a quadratic potential?

In a quadratic potential, the amplitude of oscillation remains constant as long as the restoring force remains constant. However, if the potential energy function changes, the amplitude of oscillation may also change. For example, increasing the spring constant in a mass-spring system will result in a smaller amplitude of oscillation.

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