Question 1.9: A one-dimensional harmonic oscillator of mass M and spring c...

a) The mechanical energy of the damped harmonic oscillator of a point mass M and spring constant k subjected to a damping coefficient γ, where γ \ll ω_0 and ω^2_0 = k/M , is the sum of its kinetic energy T (t) and the elastic potential energy V_e (t) of the spring,

E(t) = T(t) + V_e (t) = \frac{1}{2} M \nu ^2(t) +\frac{1}{2} kr^2(t).

where \nu ^2(t) = \dot{x} ^2(t) is the velocity squared and r^2 (t) = x^2 (t) is the displacement squared along the axis Ox. The rest position of the point mass is at the origin. Thus,

E(t) = \frac{1}{2} M \dot{x}^2(t) +\frac{1}{2} k x^2 (t).

Since the position coordinate x (t) reads,

x (t) = Ce^{−γt} \cos (ω_0t + \phi).

we have,

x^2(t) = C^2 e^{-2\gamma t} \cos ^2(\omega _0 t+\phi ).

\dot{x}^2 (t) = \omega ^{2}_{0} C^2 e^{-2\gamma t} \sin ^2(\omega _0 t+ \phi ) = \frac{k}{M} C^2 e^{-2\gamma t} \sin ^2(\omega _0 t + \phi ) .

Hence, the mechanical energy is recast as,

E(t) =\frac{1}{2} k C^2 e^{-2\gamma t}\Bigl(\sin ^2(\omega _0t + \phi )+\cos ^2(\omega _0t + \phi )\Bigr) =\frac{1}{2} kC^2 e^{-2\gamma t}.

b) The power dissipated by the friction force F_ƒ (t) is equal to the time derivative of the mechanical energy,

P(t) =\dot{E} (t) \frac{d}{dt} = \Bigl(\frac{1}{2} k C^2 e ^{-2\gamma t}\Bigr) = – \gamma k C^2 e^{-2\gamma t}.