Oscillation of bead with gravitating masses

However, at the point where the potential energy function reaches its maximum, ##U=mR##. So the period of oscillation is twice the length of the string.f
  • #1
Homework Statement
Oscillation of bead with gravitating masses:
A bead of mass m slides without friction on a smooth rod along
the x axis. The rod is equidistant between two spheres of mass M.
The spheres are located at x = 0, y = ± a as shown, and attract the
bead gravitationally.
Find the frequency of small oscillations of the bead about the
Relevant Equations
U = m * g * h
U = -g * m * M (1/r1 + 1/r2)
U = (-2 * g * m * M) / sqrt (a^2 + x^2)
The relevant equations has been me working out the gravitational potential energy. I was told to take the derivative twice from here, but I do not understand why. It leads into a taylor series expansion, which seems excessive, but I was not informed on any other way to do it. Any advice would be greatly appreciated!
  • #2
Please post the Taylor series expansion of the potential energy about ##x=0##. It is not excessive. Just show it to second order. Then we'll talk about it and you will the reason for doing it. That's my advice.

On edit: Please use ##G## instead of ##g## in your expressions. Do you see why?
  • #3
Taking the Taylor series expansion expansion about x = 0 led me to:
-(2 * G * m * M) / a + 1/2 * (2 * G * m * M) / a^3 * x^2
(note - is there a way to show fractions in a more viewable format?)

I also see why I should use G now, I had grown so accustomed to the formula the thought hadn't crossed my mind. Thank you for the correction!
I am realizing the second Taylor series expansion looks like the formula for a spring. I still don't know why it was decided to take the second derivative and use the Taylor series expansion - if I was asked to do this problem myself without any guidance, I don't know how I would arrive to the conclusion that those steps are necessary. What's a good way to think about why the solution is being approached in this way?

Edit since I wasn't super clear: The answer seems clear now. I would just not be able to replicate this process since I do not know why the second derivative is taken or what it represents, and same with the Taylor series expansion.
  • #4
The potential is an even function, meaning that ##U(-x)=U(x)##. In other words, it it symmetric about the x-axis. Furthermore, because its first derivative vanishes at x = 0, the potential has an extremum (maximum or minimum) at x = 0. Finally, the second derivative is positive, which means that you have a minimum. Thus, for small displacements from x = 0, either to the left or to the right, you get a restoring force that is proportional to the displacement just like a spring-mass system as you already noticed.

So here is the big picture. A Taylor series of an even function (a) has a first order term that vanishes; (b) has a local minimum if the second derivative at the point where the first derivative vanishes is positive. So if you find an even function expressing a potential energy, as in this problem, Taylor expand and see that it has a positive second derivative at ##x_0## where its first derivative vanishes, then you know you have simple harmonic motion about ##x_0##. Furthermore, the frequency of oscillations is related to the value of the second derivative at ##x_0##.
  • #5
I see - thank you for the help! I understand what you mean the more I read it and apply it to the problem. I have just recently learned Taylor series, so I was a bit confused on its applications. I have a more solid grasp on the derivative part of your explanation, but I'll likely understand Taylor series the more I use them. Thank you again!
  • #6
You might wish to practice by considering the following application. You have a bead of mass ##m## that slides without friction on a ring of radius ##R##. The ring is oriented so that it has a diameter along the vertical direction, parallel to the acceleration of gravity. Find the period of small oscillations of the mass about the lowest point of the ring.

Comment 1
The motion of the mass is circular which is the same as if it were attached to a string to form a pendulum. You expect the period to be the same as that of a pendulum of length ##R##. This application shows you how to get the answer using Taylor series without recourse to Newtons's second law.

Comment 2
The equation of a circle is ##x^2+y^2=R^2.##
The potential energy function is ##U=mgy## and increases as ##y## increases.

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