Damped Simple Harmonic Motion — Question 1

Question 1

A decay record separates damping from stiffness

A mass $m=.50\,\mathrm{kg}$ obeys $m\ddot x+b\dot x+kx=0$. Successive positive displacement maxima are separated by $T_d=.80\,\mathrm{s}$ and have ratio $x_n/x_{n+1}=1.50$. Assume linear viscous damping.

1. Infer the decay constant $\beta=b/(2m)$, damping coefficient $b$, and spring constant $k$.

2. Compare the damped and undamped periods. Why does substituting $T_d$ directly into $k=m(2\pi/T)^2$ underestimate stiffness?

Question 1 - Solution

A decay record separates damping from stiffness

(a) Derivation and calculation

$$\omega_d=2\pi/T_d=7.85398\,\mathrm{rad/s},\quad\frac{x_n}{x_{n+1}}=e^{\beta T_d}.$$$$\boxed{\beta=\ln(1.50)/.800=.506831\,\mathrm{s^{-1}}},\quad\boxed{b=2m\beta=.506831\,\mathrm{kg/s}}.$$$$\omega_0^2=\omega_d^2+\beta^2\Rightarrow\boxed{k=m(\omega_d^2+\beta^2)=30.9709\,\mathrm{N/m}}.$$

(b) Result and interpretation

$$\omega_0=7.87032\,\mathrm{rad/s},\qquad\boxed{T_0=2\pi/\omega_0=.79834\,\mathrm s<T_d}.$$$$k_{\mathrm{naive}}=m\omega_d^2=30.8425\,\mathrm{N/m},\quad\boxed{k-k_{\mathrm{naive}}=m\beta^2=.12844\,\mathrm{N/m}}.$$Peak spacing measures the damped frequency; stiffness determines the undamped frequency.

Physical check

The correction is positive and tends to zero as the measured peak ratio tends to one.