What Is the Frequency of Oscillation When a Mass Suspends from a Spring?

[RJLiberator]

There is a part c and part d of this problem.
I'm almost at my limit, but I felt like posting this here if you guys wanted to review my work.
If you want, you can just reply "Wrong or Right," and I'd be happy. If not, no bothers, I didn't feel like this deserved it's own thread.

A second mass of 0.3 kg is added to the first mass, making a total of M+0.3kg. When this system oscillates, it has half the frequency of the system with mass M alone. (c) What is the value of M? (d) Where is the new equilibrium position?

c) I use a relationship
w^2 = k/M from original problem in this thread
and (1/2*w)^2 = k/(M+0.3)

With a little algebraic manipulation we see 3M = 0.3 so M = 0.1 kg is the answer.

d) I use w^2 = k/M from original problem and our new mass M = 0.1 to find spring constant K.
K = 19.628.
Now, I use this information and apply it to the equation to find amplitude K = mg/x where m is now 0.3+0.1 = 0.4 and x is what we are looking for.
Solving this gives me x = 0.2 from the initial point.

Using my knowledge I learned in this thread, I know this is 0.4m from the ceiling and so the new equilibrium position is now 0.3m.I feel confident with these answers and my strategy to tackle them, but as this thread has shown, my understanding of classical mechanics is tarnished :).
Are my strategies correct?

 

[haruspex]

I agree with your answers, except that I don't understand the final conversion to .3m in d. Where are you measuring it from to get that?

You could get d a little more easily. If the equilibrium extension for the original mass was .05m, the eq. ext. for four times the mass must be four times that.

 

[RJLiberator]

I get .3m from looking at the initial experiment.

The initial parts a and b saw that when the mass went 0.1 meters beneath the 0.2m initial starting point, that the equilibrium point was 1/2 of 0.1 m at 0.05m lower than 0.2m.
So I applied that same practice when the mass dropped the string 0.2 meters more to 0.4meters. So the oscillation is from 0.4 meter mark to 0.2 meter mark and therefore the 0.3 mark must be the equilibrium :).

I'm glad that I am starting to get this problem. It's clear to me that I've learned (something) in this thread.

 

[haruspex]

So I applied that same practice when the mass dropped the string 0.2 meters more to 0.4meters. So the oscillation is from 0.4 meter mark to 0.2 meter mark and therefore the 0.3 mark must be the equilibrium :).

But the .2m you calculated was from zero extension to the new equilibrium (4x0.05), not to the new lowest point. The new lowest point would be .4m from zero extension, making it a total length of .6m.