Simple Harmonic Oscillator Troubles

[DougD720]

Homework Statement

This is a 3 part problem, mass M on a spring of length l with mass m. The first part was to derive the Kinetic Energy of one segment dy, second part was to Integrate this and get the Kinetic Energy of (1/6)m(V^2) where V is the velocity of the Mass M at the end of the spring oscillating.

The final part of the problem and the part which I've been trying to work out for too long now, is to prove via the equation for total energy, that:

w^2 = s / (M + m/3)

Which I keep getting very close to but cannot get exactly.

Homework Equations

E = KE + PE = (Constant) => E = (1/6)m(x')^2 + (1/2)sx^2

dE/dt = 0 = (1/3)mx" + sx

x = a sin(wt)
x' = aw cos(wt)
x" = -a(w^2) sin(wt)

The Attempt at a Solution

I keep getting the following 2 equations:

w^2 = s / (m/3)

E = (a^2 w^2)(m/6 cos^2(wt) + M/2 sin^2(wt))

I feel like I'm making a stupid mistake or something because I keep trying these equations different ways with different trig identities, etc. but keep getting the wrong answer. The first equation there is close, but if I'm supposed to prove the value of w^2, I can't incorporate the equation w^2 = s/M and if I can't use that how do I get M into that equation?

I'm either missing something here or making a mistake in my logic, any help would be greatly appreciated!

Thanks!

 

[Donaldos]

The expression for total energy is:

$$E=\left(\frac M 2 + \frac 1 6 m\right)v^2+\frac 1 2 s x^2 + Mgx$$

$$\frac{{\rm d} E}{{\rm d} t}=0 \Rightarrow v\left[\left( M + \frac 1 3 m\right)\frac{{\rm d}^2 x}{{\rm d} t^2}+s x + Mg\right]=0$$

$$\left( M + \frac 1 3 m\right)\frac{{\rm d}^2 x}{{\rm d} t^2}+s x = -Mg$$

To be continued...

 

[DougD720]

Ah... Think I've got it now, I wasn't adding in the KE of the Mass on the End, 1/2 M V^2, I was under the (incorrect) impression it was bundled in with the (1/6)mV^2... knew it was a dumb mistake on my part.

Thanks!