Differential Equation Mass Damped Spring I have already attempted to find the natural frequency by taking the root of $k/\mu$ where $\mu = 2m$. I then divided that by $2\pi$ to get $6.74\cdot 10^{13}$, not sure how to answer the second question though.


*

*The $\mu$ for the Oxygen molecule (O2) is $1.33\cdot 10^{-26}$kg and $k =1195$N/m. What is the natural frequency of O2?

*What will happen to the molecule if it is forced by an external source to vibrate with a 
frequency equal to its natural frequency? Explain in detail.
 A: In particular, if $y$ is the distance from the equilibrium position, the equation (in the linear realm) for $y$ if the forcing has the same frequency as the natural frequency is
$$
\frac{d^2y}{dt^2} + \frac{k}{\mu}y -a\cos \left( \sqrt{\frac{k}{\mu}} t\right) = 0
$$
and this is solved (starting at rest at $y=0$ by 
$$
y = \frac{a}{2}\sqrt{\frac{\mu}{k}} t \sin  \left( \sqrt{\frac{k}{\mu}} t\right) 
$$
so the amplitude of vibration grows linearly with time.  
Of course, in a real molecule the restoring force becomes non-linear way before the molecule dissociates, but you can get a good estimate of the time to dissociation by seeing when the energy given by that formula exceeds the molecular binding energy.
A: If you force a spring using its natural frequency the amplitude will get larger and larger. Obviously the linear spring model will break down at some point, in particular, the $O_2$ molecule will dissociate into two oxygen atoms.
BY the way, you might want to check your math for part 1.  Did you use the reduced mass $\mu$ or the mass of one oxygen atom?
A: If the molecule is being forced to vibrate at its natural frequency, my guess is that the molecule would break apart. I am not a chemist so I am not entirely certain. I am saying based on what happens to structures that are forced to vibrate at their natural frequency. For instance, take the Tacoma National Bridge which you can watch here on Youtube.
Tacoma Bridge
