# Oscillation of bead with gravitating masses

• Penny57
In summary: 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.
Penny57
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
origin.
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!

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.

Penny57
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.

knightlyghost and haruspex
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##.

Penny57
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!

kuruman
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.

## 1. What is oscillation of bead with gravitating masses?

Oscillation of bead with gravitating masses refers to the motion of a small bead on a wire or string that is affected by the gravitational pull of two or more larger masses.

## 2. How does the mass of the bead affect its oscillation?

The mass of the bead does not significantly affect its oscillation, as long as it is much smaller than the gravitating masses. The motion of the bead is primarily determined by the gravitational forces acting on it.

## 3. What factors affect the period of oscillation for the bead?

The period of oscillation for the bead is affected by the distance between the two gravitating masses, the masses of the gravitating objects, and the length of the wire or string that the bead is suspended from.

## 4. How is the oscillation of the bead affected by the strength of gravity?

The strength of gravity does not significantly affect the oscillation of the bead, as long as the distances between the bead and the gravitating masses are much larger than the size of the bead itself.

## 5. Can the oscillation of the bead be used to measure the masses of the gravitating objects?

Yes, the period of oscillation of the bead can be used to calculate the masses of the gravitating objects using Newton's Law of Universal Gravitation. However, this method may not be very accurate as it assumes the bead is a point mass and does not take into account other factors such as air resistance or the shape of the gravitating masses.

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