One spring, two masses, on a frictionless surface

[arcraith]

Homework Statement

A mass is hung from a spring which is secured to the ceiling. We measure the frequency of the oscillation of the mass. Now we attach a second identical mass to the other end of the spring and set the system on a frictionless surface. If we now compress the spring and release the ends, what frequency of oscillation do you expect to measure? Why?

Homework Equations

f = (1/(2pi))(k/m)^(1/2)
F = -ky
E = (1/2)kA^2

The Attempt at a Solution

Apparently the frequency of the second system is:

f = (1/(2pi))(2k/m)^(1/2)
NOTE THE "2k" over m

I've been stuck on this for ages, and i still don't know how to get this answer.

the spring constant shouldn't change since its the same spring, so the difference must have to do with the added mass?

i've tried thinking of ways to model the second system as a spring with one mass so i can obtain a ratio of the 'apparent' masses or spring constants, but all i can come up with is thinking of the middle of the spring as being the ceiling like in the first system, but that gives me a spring with the same constant and half as long, and doesn't work since the frequency doesn't depend on the length.

and beyond that I'm completely stumped. i really need to understand how this works.

thanks in advance.

 

[anathema]

Hello, thank you for your post. I understand your confusion and I will do my best to explain the solution.

First, let's review the equations that you have listed. The first equation, f = (1/(2pi))(k/m)^(1/2), is the equation for the frequency of an oscillating mass attached to a spring, where k is the spring constant and m is the mass. This equation assumes that there is only one mass attached to the spring. The second equation, F = -ky, is the equation for the force exerted by a spring, where k is the spring constant and y is the displacement from the equilibrium position. The third equation, E = (1/2)kA^2, is the equation for the potential energy of a spring, where k is the spring constant and A is the amplitude of the oscillation.

Now, let's apply these equations to the first system. In the first system, there is only one mass attached to the spring, so we can use the first equation to calculate the frequency. If we add a second identical mass to the other end of the spring, the total mass attached to the spring will be 2m. However, the spring constant will remain the same, as it is a property of the spring itself and does not change unless the spring is physically altered. This means that the frequency of the second system will be f = (1/(2pi))(2k/2m)^(1/2) = (1/(2pi))(k/m)^(1/2), which is the same as the frequency of the first system.

Now, let's consider the situation where the system is placed on a frictionless surface and the spring is compressed and released. When the spring is compressed, the potential energy of the system is increased. As the spring is released, this potential energy is converted into kinetic energy, causing the mass to oscillate. The amplitude of this oscillation will depend on the amount of potential energy that was stored in the spring. In this case, the potential energy is given by E = (1/2)kA^2, where A is the amplitude of the oscillation. Since the spring constant remains the same, the amplitude will also remain the same for both systems.

Now, let's put all of this together. The frequency of an oscillating mass attached to a spring is given by f = (1/(2pi))(k/m