Q: A mass $m$ hangs from a massless spring connected to the roof of a box of mass $M$. When the box is held stationary, the mass-spring system oscillates vertically with angular frequency $\omega$. If the box is dropped falls freely under gravity, why does the angular frequency increase?
So initially, we have $\omega=\sqrt{\frac{k}{m}}$, where $k$ is the spring constant. I read that the new angular frequency when in free-fall is $\omega'=\sqrt{\frac{k}{\mu}}$, where $\mu=\frac{Mm}{M+m}$ is the reduced mass, i.e. $\omega'=\sqrt{\frac{k(M+m)}{Mm}}$ which is clearly greater than $\omega$.
I'm not sure why this is true – I suspect it has something to do with the box oscillating (as it's no longer held stationary), but I'm not too familiar with the concept of reduced mass so I'd really appreciate a good explanation.
Best Answer
Despite the weird geometry of the problem, this is simply two masses connected by a spring, and we can solve that problem using Lagrangian Mechanics. Let's assume that the masses have mass $M$ and m, and position coordinates $x_M$ and $x_m$ respectivly. They are also separated by a distance $d$, and the spring constant is $k$.
We will use $\alpha$ as the generalized coordinate representing how far the masses have stretched from equilibrium: $$ \alpha = x_M - x_m - d $$
Note also that $$ \dot{\alpha} = \dot{x}_M - \dot{x}_m, $$ because $d$ is a constant, and we can choose a frame in which the total momentum of the system is zero, which gives us $$ m \dot{x}_m = -M \dot{x}_M $$
We can write the kinetic energy as $$ T = \frac{1}{2} m \dot{x}_m^2 + \frac{1}{2} M \dot{x}_M^2, $$ and then transform to our generalized coordinate $\alpha$ using the above two equations: $$ T = \frac{1}{2} m \dot{x}_m^2 + \frac{1}{2} M \Big(\dot{x}_m \frac{m}{M}\Big)^2 $$ $$ T = \frac{1}{2} m \Big( -\dot{\alpha} \frac{M}{m+M}\Big)^2 + \frac{1}{2} M \Big( \Big( -\dot{\alpha} \frac{M}{m+M}\Big) \frac{m}{M}\Big)^2 $$ $$ T= \frac{1}{2} \frac{m M^2}{(m+M)^2} \dot{\alpha}^2+ \frac{1}{2} \frac{m^2 M}{(m+M)^2}\dot{\alpha}^2 $$ $$ T= \frac{1}{2} \frac{m M}{m+M} \dot{\alpha}^2 $$ $$ T= \frac{1}{2} \mu \dot{\alpha}^2, $$ where $\mu$ is the reduced mass.
The potential energy is much easier to find: $$ U = \frac{1}{2} k \alpha^2. $$ Our Lagrangian then is $$ L = T-U $$ $$ L = \frac{1}{2}\mu \dot{\alpha}^2 - \frac{1}{2} k \alpha^2, $$ which can be plugged into the Euler-Lagrange equation
$$ \frac{d}{dt} \frac{\partial L}{\partial \dot{\alpha}} - \frac{\partial L}{\partial \alpha}=0. $$
This equation is easy to solve by hand: $$ \mu \ddot{\alpha}+k \alpha = 0. $$
This is clearly the equation for a simple harmonic oscillator with angular frequency $\sqrt{k/\mu}$, which is the answer you were looking for!