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oscillation in quadratic potential |
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| Dec27-10, 09:28 AM | #1 |
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oscillation in quadratic potential
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. |
| Dec27-10, 02:59 PM | #2 |
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Hi Idoubt!
![]() What equations do you know relating the potential to the motion?
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| Dec27-10, 08:54 PM | #3 |
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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. |
| Dec28-10, 03:05 AM | #4 |
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oscillation in quadratic potential
How about a differential equation?
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| Dec28-10, 06:54 AM | #5 |
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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
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| Dec28-10, 08:16 AM | #6 |
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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? |
| Dec28-10, 01:44 PM | #7 |
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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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| Dec28-10, 02:31 PM | #8 |
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Mentor
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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. |
| Dec29-10, 03:16 AM | #9 |
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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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| Dec29-10, 03:29 AM | #10 |
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yes!
and yes!
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| Dec29-10, 05:12 AM | #11 |
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Recognitions:
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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.
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