Published by Patrick Mutisya · 14 days ago
Students will understand that resonance corresponds to a maximum amplitude of oscillation and occurs when a system is forced to oscillate at (or very near) its natural frequency.
For an undamped mass‑spring system:
\$\$
m\ddot{x}+kx=0
\$\$
with natural (angular) frequency
\$\$
\omega_0=\sqrt{\frac{k}{m}}.
\$\$
When a resistive force proportional to velocity, \$-b\dot{x}\$, is present, the equation becomes
\$\$
m\ddot{x}+b\dot{x}+kx=0,
\$\$
or, using the damping coefficient \$\beta = \dfrac{b}{2m}\$,
\$\$
\ddot{x}+2\beta\dot{x}+\omega_0^{2}x=0.
\$\$
| Regime | Condition | Behaviour |
|---|---|---|
| Underdamped | \$\beta < \omega_0\$ | Oscillatory motion with exponentially decaying amplitude. |
| Critically damped | \$\beta = \omega_0\$ | Returns to equilibrium as quickly as possible without overshoot. |
| Overdamped | \$\beta > \omega_0\$ | Non‑oscillatory return to equilibrium, slower than critical case. |
The displacement can be written as
\$\$
x(t)=A e^{-\beta t}\cos(\omega_d t+\phi),
\$\$
where the damped angular frequency is
\$\$
\omegad=\sqrt{\omega0^{2}-\beta^{2}}.
\$\$
The total mechanical energy decays as
\$\$
E(t)=E_0 e^{-2\beta t}.
\$\$
Adding a sinusoidal driving force \$F(t)=F_0\cos(\omega t)\$ gives
\$\$
m\ddot{x}+b\dot{x}+kx = F_0\cos(\omega t).
\$\$
After transients die out, the displacement is
\$\$
x(t)=A(\omega)\cos(\omega t-\delta),
\$\$
with amplitude
\$\$
A(\omega)=\frac{F0/m}{\sqrt{(\omega0^{2}-\omega^{2})^{2}+(2\beta\omega)^{2}}}
\$\$
and phase lag
\$\$
\tan\delta=\frac{2\beta\omega}{\omega_0^{2}-\omega^{2}}.
\$\$
Resonance occurs when the driving frequency \$\omega\$ is such that the amplitude \$A(\omega)\$ reaches a maximum.
Setting \$\dfrac{dA}{d\omega}=0\$ yields the resonance (angular) frequency
\$\$
\omega{\text{res}}=\sqrt{\omega0^{2}-2\beta^{2}}.
\$\$
For light damping (\$\beta\ll\omega0\$) this reduces to \$\omega{\text{res}}\approx\omega_0\$, i.e. the natural frequency.
The sharpness of the resonance is characterised by the quality factor
\$\$
Q=\frac{\omega_0}{2\beta}=\frac{1}{2\beta}\sqrt{\frac{k}{m}}.
\$\$
A high \$Q\$ means a narrow, high peak in the amplitude–frequency curve.
At resonance the work done by the driving force per cycle is maximal, because the force and velocity are in phase (phase lag \$\delta\approx\pi/2\$).
Problem: A mass \$m=0.5\;\text{kg}\$ attached to a spring of constant \$k=200\;\text{N m}^{-1}\$ is driven by a force \$F(t)=10\cos(15t)\;\text{N}\$. The damping constant is \$b=2\;\text{kg s}^{-1}\$. Determine the amplitude of the steady‑state motion and state whether the system is near resonance.
Solution Sketch:
\$A=\frac{F0/m}{\sqrt{(\omega0^{2}-\omega^{2})^{2}+(2\beta\omega)^{2}}},\$
where \$F_0=10\;\text{N}\$, \$\omega=15\;\text{rad s}^{-1}\$.
\$\$A=\frac{10/0.5}{\sqrt{(20^{2}-15^{2})^{2}+(2\times2\times15)^{2}}}
=\frac{20}{\sqrt{(400-225)^{2}+ (60)^{2}}}
=\frac{20}{\sqrt{175^{2}+60^{2}}}
\approx\frac{20}{185}\approx0.108\;\text{m}.\$\$