Damped Simple Harmonic Motion — Question 4

Question 4

Adjacent extrema and same-side peaks imply different decrements

The absolute displacements at successive extrema of an underdamped viscous oscillator are $12.0$, $9.0$ and $6.75\,\mathrm{mm}$. Adjacent extrema are $.20\,\mathrm{s}$ apart. Their signs alternate.

1. Find $T_d$, $\beta$ and the logarithmic decrement between successive positive maxima.

2. Find the mechanical-energy ratio between adjacent extrema. Explain the factor-of-two mistake made by treating adjacent extrema as a full period.

Question 4 - Solution

Adjacent extrema and same-side peaks imply different decrements

(a) Derivation and calculation

$$\Delta t=T_d/2=.200\,\mathrm s\Rightarrow\boxed{T_d=.400\,\mathrm s}.$$$$e^{-\beta\Delta t}=9/12=.750\Rightarrow\boxed{\beta=\frac{\ln(4/3)}{.200}=1.43841\,\mathrm{s^{-1}}}.$$$$\boxed{\delta=\beta T_d=2\ln(4/3)=.575364}.$$

(b) Result and interpretation

At an extremum, $E=kx^2/2$:$$\boxed{\frac{E_{n+1}}{E_n}=\left(\frac{|x_{n+1}|}{|x_n|}\right)^2=.750^2=.5625}.$$$$\omega_d=2\pi/.400=15.708\,\mathrm{rad/s}.$$Treating $.200\,\mathrm s$ as a full period doubles the inferred frequency; using $\ln(4/3)$ as the same-side decrement halves its true value.

Physical check

The two-step ratio is $6.75/12=.5625=e^{-\delta}$, confirming the same-side decrement.