Physics Asked on June 22, 2021
Suppose you have an harmonic oscillator whose mass is adiabatically changing such that $Tfrac{dm}{dt}ll m$ where $T$ is the period of the motion. It could for example be an ice ball slowly melting connected to a spring.
Considering the adiabatic invariant $frac{E}{omega}$ where $E$ is the energy of the oscillator and $omega^2=frac{k}{m}$, the energy of the oscillator should increase. However I can’t figure out why this happens by starting from first principles (like Newton equations). How does it work?
The equation of motion for " ice ball slowly melting connected to a spring"
you start with the kinetic energy $T_c$
$$T_c=frac 12 m(tau),dot x^2$$
and the potential energy of the spring $U$
$$U=frac 12 k,x^2$$
with Euler- Lagrange you obtain the EOM.
$$ begin {array}{c} m left( tau right) {ddot x}+kx+ left( { frac {d}{dtau}}m left( tau right) right) {dot x}end {array} =0 $$
you can make this Ansatz for $m(tau)$
$$m(tau)=m_{{0}},{{e}^{-{frac {tau}{T}}}}$$
where $T$ is the time constant $~[s]$ for the ice melting process and $m_0$ is the initial ice mass
the EOM is now
$$ddot x+underbrace{frac{e^{(tau/T)},k}{m_0}},x-frac{1}{T},dot x$$
Edit
The total energy is:
$$E=frac 12,m_{{0}}{{rm e}^{-{frac {tau}{T}}}}{{dot x}}^{2}+frac 12,k{x}^{2}$$
with :
$$T=frac {2pi}{omega}=frac{2,pi}{sqrt{frac{k}{m_0}}}~,k=m_0,omega^2$$
$$E=frac 12,m_{{0}}{{rm e}^{-frac{1}{2,pi},sqrt {{frac {k}{m_{{0}}}}}tau}}{{dot x}}^{2}+frac 12,m_{{0}}{omega}^{2}{x}^{2} $$
for $tau mapsto infty$
$$E=frac 12,m_{{0}}{omega}^{2}{x}^{2} ~,frac{E}{omega}=frac 12,m_0,omega,x^2=text{const.}$$
Answered by Eli on June 22, 2021
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