Forced Oscillations and Resonance — Question 9

Question 9

Energy supplied per cycle is determined by the force-displacement loop

In steady driven motion, $x=X\cos(\Omega t-\delta)$ and the applied force is $F=F_0\cos\Omega t$. Let $\Omega>0$ and $0<\delta<\pi$.

1. Evaluate the work supplied in one cycle and relate it to the oriented force-displacement loop.

2. For $F_0=3.0\,\mathrm{N}$, $X=.020\,\mathrm{m}$, $\delta=\pi/6$ and $\Omega=10\,\mathrm{rad/s}$, find the work per cycle, mean power and compatible viscous coefficient $b$.

Question 9 - Solution

Energy supplied per cycle is determined by the force-displacement loop

(a) Derivation and calculation

Put $\psi=\Omega t-\delta$; then $F=F_0\cos(\psi+\delta)$ and $dx=-X\sin\psi\,d\psi$.$$W=-F_0X\int_0^{2\pi}(\cos\psi\cos\delta-\sin\psi\sin\delta)\sin\psi\,d\psi.$$$$\boxed{W=\oint F\,dx=\pi F_0X\sin\delta=\pi b\Omega X^2}.$$

(b) Result and interpretation

$$\boxed{W=\pi(3.00)(.0200)\sin(\pi/6)=.094248\,\mathrm J}.$$$$\boxed{\overline P=\frac{W\Omega}{2\pi}=.150\,\mathrm W}.$$$$\boxed{b=\frac{F_0\sin\delta}{\Omega X}=\frac{3.00(.500)}{10(.0200)}=7.50\,\mathrm{kg/s}}.$$The loop integral is positive and replaces the energy dissipated during a cycle.

Physical check

Stored mechanical energy returns to its original value each steady cycle, so net input equals damping loss.