Exercise 3.5: Resonance

The long term behavior of the driven harmonic oscillator depends on the frequency of the driving force. One measure of this behavior is the maximum of the steady state displacement $A(\omega )$.

1. Adopt the initial condition $(x_0=0,v_0=0)$. Compute $A(\omega )$ for $\omega=0$, $1.0$, $2.0$, $2.2$, $2.4$, $2.6$, $2.8$, $3.0$, $3.2$, $3.4$ with $\omega _0=3$ and $\gamma=0.5$. Plot $A(\omega )$ versus $\omega$ and describe its qualitative behavior. If $A(\omega )$ has a maximum, determine the ``resonance angular frequency'' $\omega _{max}$, which is the frequency at the maximum of $A$. Is the value of $\omega _{max}$ close to the natural angular frequency $\omega _0$?

2. Compute $A_{max}$, the value of the amplitude at $\omega _{max}$, and the ratio $\Delta \omega /\omega _{max}$, where $\Delta \omega$ is the ``width'' of the resonance. Define $\Delta \omega$ as the frequency interval between points on the amplitude curve which are $1/\sqrt{2}A_{max}$. Set $\omega _0=3$ and consider $\gamma=0.1$, 0.5, 1.0, 2.0. Describe the qualitative dependence of $A_{max}$ and $\Delta \omega /\omega _{max}$ on $\gamma$. The quantity $\Delta \omega /\omega _{max}$ is proportional to $1/Q$, where $Q$ is the ``quality factor'' of the oscillator.

3. Decribe the qualitative behavior of the steady state amplitude $A(\omega )$ near $\omega=0$ and $\omega\gg \omega _0$. Why is $A(\omega =0) > A(\omega )$ for small $\omega$? Why does $A(\omega ) \rightarrow 0$ for $\omega\gg \omega _0$?