Students will be able to:
For an ideal (undamped) mass‑spring system
\[
m\ddot{x}+kx=0
\]
the natural (angular) frequency and ordinary frequency are
\[
\omega{0}=\sqrt{\frac{k}{m}}, \qquad f{0}=\frac{\omega_{0}}{2\pi}
\]
When a resistive force proportional to velocity, \(F_{\text d}=-b\dot{x}\), acts, the equation becomes
\[
m\ddot{x}+b\dot{x}+kx=0
\]
Introduce the damping coefficient
\[
\beta=\frac{b}{2m}
\]
and the dimensionless damping ratio
\[
\zeta=\frac{\beta}{\omega_{0}}=\frac{b}{2\sqrt{km}}
\]
| Regime | Condition (using \(b\) or \(\zeta\)) | Typical behaviour |
|---|---|---|
| Light (underdamped) | \(b<2\sqrt{km}\) or \(\zeta<1\) | Oscillatory motion with an exponentially decaying envelope. |
| Critical | \(b=2\sqrt{km}\) or \(\zeta=1\) | Returns to equilibrium as quickly as possible without overshoot; the envelope decays as fast as possible. |
| Heavy (overdamped) | \(b>2\sqrt{km}\) or \(\zeta>1\) | No oscillation; the system slowly returns to equilibrium via two real exponentials. |
\[
x(t)=A\,e^{-\beta t}\cos(\omega_{d}t+\phi),\qquad
\omega{d}=\sqrt{\omega{0}^{2}-\beta^{2}}
\]
\[
x(t)=(A+Bt)\,e^{-\omega_{0}t}
\]
Note that the condition can also be written as \(b=2\sqrt{km}\).
\[
x(t)=C\,e^{-(\beta+\sqrt{\beta^{2}-\omega_{0}^{2}})t}
+D\,e^{-(\beta-\sqrt{\beta^{2}-\omega_{0}^{2}})t}
\]
Both terms are real exponentials; because the exponents are negative, the motion never changes sign – no oscillation occurs.
The total mechanical energy falls exponentially:
\[
E(t)=E_{0}\,e^{-2\beta t}
\]
\[
m\ddot{x}+b\dot{x}+kx = F_{0}\cos(\omega t)
\]
The solution consists of
After a few periods the motion is essentially
\[
x(t)=A(\omega)\cos\!\bigl(\omega t-\delta\bigr)
\]
\[
A(\omega)=\frac{F_{0}/m}
{\sqrt{\bigl(\omega_{0}^{2}-\omega^{2}\bigr)^{2}
+\bigl(2\beta\omega\bigr)^{2}}}
\]
\[
\tan\delta=\frac{2\beta\omega}{\;\omega_{0}^{2}-\omega^{2}\;}
\]
Maximum amplitude occurs when \(\dfrac{dA}{d\omega}=0\). It is algebraically simpler to differentiate \(A^{-2}\):
\[
\frac{d}{d\omega}\Big[(\omega_{0}^{2}-\omega^{2})^{2}
+(2\beta\omega)^{2}\Big]=0
\]
\[
-4\omega(\omega_{0}^{2}-\omega^{2})+8\beta^{2}\omega=0
\]
\[
\boxed{\;\omega{\text{res}}^{2}= \omega{0}^{2}-2\beta^{2}\;}
\]
For light damping (\(\beta\ll\omega{0}\)) the resonance angular frequency is essentially the natural frequency, \(\omega{\text{res}}\approx\omega_{0}\).
Resonance is the condition in which a driven oscillator exhibits the largest possible steady‑state amplitude for a given driving force.
The plot of \(A(\omega)\) versus \(\omega\) shows a peak at \(\omega_{\text{res}}\). The sharpness of the peak depends on the damping:
Quality factor (how “selective’’ the resonance is)
\[
Q=\frac{\omega_{0}}{2\beta}=\frac{1}{2\zeta}
\]
Bandwidth (full‑width at half‑maximum) for a lightly damped oscillator
\[
\Delta\omega\;\approx\;\frac{\omega_{0}}{Q}
\]
Interpretation:
The instantaneous power supplied by the driver is \(P=F(t)v(t)\). At resonance the phase lag \(\delta\approx\pi/2\), so force and velocity are in phase, giving the maximum average power per cycle. Consequently the mechanical energy stored in the oscillator grows until the damping losses balance the input power.
Problem: A mass \(m=0.5\;\text{kg}\) is attached to a spring of constant \(k=200\;\text{N m}^{-1}\). The system is driven by a force \(F(t)=10\cos(15t)\;\text{N}\). The damping constant is \(b=2\;\text{kg s}^{-1}\). Determine:
Solution:
\[
\omega_{0}= \sqrt{\frac{k}{m}}
=\sqrt{\frac{200}{0.5}}=20\;\text{rad s}^{-1}
\]
Damping coefficient and ratio
\[
\beta=\frac{b}{2m}= \frac{2}{2\times0.5}=2\;\text{s}^{-1},
\qquad \zeta=\frac{\beta}{\omega_{0}}=0.10\;( \text{light damping})
\]
Amplitude
\[
A=\frac{F_{0}/m}
{\sqrt{(\omega_{0}^{2}-\omega^{2})^{2}
+(2\beta\omega)^{2}}}
=\frac{10/0.5}
{\sqrt{(20^{2}-15^{2})^{2}+(2\times2\times15)^{2}}}
=\frac{20}{\sqrt{175^{2}+60^{2}}}
\approx 0.108\;\text{m}
\]
\[
\omega{\text{res}}=\sqrt{\omega{0}^{2}-2\beta^{2}}
=\sqrt{400-8}= \sqrt{392}\approx19.8\;\text{rad s}^{-1}
\]
The driving frequency \(\omega=15\;\text{rad s}^{-1}\) is well below \(\omega_{\text{res}}\); the system is not near resonance, which explains the relatively modest amplitude.
\[
Q=\frac{\omega_{0}}{2\beta}= \frac{20}{4}=5
\]
\[
\Delta\omega\approx\frac{\omega_{0}}{Q}= \frac{20}{5}=4\;\text{rad s}^{-1}
\]
The half‑power points lie roughly at \(\omega_{\text{res}}\pm\frac{\Delta\omega}{2}\approx19.8\pm2\;\text{rad s}^{-1}\). The driving frequency \(15\;\text{rad s}^{-1}\) lies far outside this resonant band.
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