Physics
Damped Oscillators types mathematical function
Step-by-step physics solution: Damped Oscillators types mathematical function
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1. What the student is asking
The student has three questions:
-
Why does the textbook formula
[ y(t)=y_0\,e^{-\delta t}\cos (\omega t) ]
only describe one kind of damping?
-
What are the correct mathematical conditions for under‑, critical‑ and overdamping? (The teacher said “critical damping when δ = ω and overdamping when δ > ω”, but the student’s simulations disagree.)
-
Why does a larger exponential decay constant δ sometimes make the motion slower (over‑damped case)?
Below we answer each point step‑by‑step, starting from the differential equation that governs a damped harmonic oscillator.
2. Derivation from the equation of motion
A mass‑spring system with a viscous damper obeys
[ m\ddot y + b\dot y + k y = 0, ]
where
- (m) = mass,
- (b) = damping coefficient (units N·s m(^{-1})),
- (k) = spring constant.
It is convenient to define two angular frequencies:
[ \boxed{\omega_0 \equiv \sqrt{\frac{k}{m}}}\qquad\text{(undamped natural frequency)} ]
[ \boxed{\delta \equiv \frac{b}{2m}}\qquad\text{(damping constant, sometimes called “decay rate”)}. ]
Dividing the whole equation by (m) gives the standard form
[ \ddot y + 2\delta \dot y + \omega_0^{\,2} y = 0. \tag{1} ]
Equation (1) is a second‑order linear ODE with constant coefficients.
Assume a solution of the form (y(t)=e^{\lambda t}). Substituting yields the characteristic equation
[ \lambda^{2}+2\delta\lambda+\omega_0^{2}=0. ]
Its roots are
[ \boxed{\lambda_{1,2}= -\delta \pm \sqrt{\delta^{2}-\omega_0^{2}} } .\tag{2} ]
The nature of the square‑root term determines the type of damping.
3. Three regimes and their explicit solutions
| Regime | Condition on (\delta) and (\omega_0) | Roots (\lambda_{1,2}) | General solution (y(t)) | Behaviour |
|---|---|---|---|---|
| Underdamped | (\displaystyle \delta < \omega_0) | (\lambda = -\delta \pm i\omega_D) with (\displaystyle\omega_D =\sqrt{\omega_0^{2}-\delta^{2}}) (purely imaginary part) | (\displaystyle y(t)=e^{-\delta t}\big(A\cos\omega_D t + B\sin\omega_D t\big)) or (y(t)=y_0 e^{-\delta t}\cos(\omega_D t+\phi)) |
Oscillatory with exponentially decreasing envelope. |
| Critically damped | (\displaystyle \delta = \omega_0) | (\lambda = -\delta) (double root) | (\displaystyle y(t)=\big(A+Bt\big)\,e^{-\delta t}) | Returns to equilibrium as fast as possible without oscillating. |
| Overdamped | (\displaystyle \delta > \omega_0) | (\lambda_{1,2}= -\delta \pm \sqrt{\delta^{2}-\omega_0^{2}}) – both are real and negative | (\displaystyle y(t)=A\,e^{\lambda_1 t}+B\,e^{\lambda_2 t}) with (\lambda_1\neq\lambda_2) | No oscillation; the motion is a sum of two decaying exponentials. The slower‑decaying term dominates at long times, giving a “gliding” approach to equilibrium. |
Why the teacher wrote “δ = ω”?
Often textbooks rename the undamped natural frequency (\omega_0) simply as (\omega).
If the teacher used that shorthand, the statement “critical damping when (\delta = \omega)” really means
[ \boxed{\delta = \omega_0}. ]
The student’s confusion arises because the formula they were given, (y(t)=y_0 e^{-\delta t}\cos(\omega t)), already assumes the under‑damped case and uses
[ \omega = \omega_D = \sqrt{\omega_0^{2}-\delta^{2}} . ]
Thus the same symbols appear in two different contexts, which is why the simulation did not match the teacher’s rule.
4. Visualising the three cases
Below are the typical shapes (choose arbitrary (y_0=1), (m=1) for simplicity).
| Parameter set | Plot description |
|---|---|
| Underdamped: (\omega_0 = 5\; \text{rad/s},\; \delta = 1\; \text{s}^{-1}) | Oscillations with period (2\pi/\omega_D \approx 1.3) s, envelope (e^{-t}). |
| Critical: (\omega_0 = \delta = 5\; \text{s}^{-1}) | Curve rises (or falls) monotonic, reaches zero fastest among all non‑oscillatory curves. |
| Overdamped: (\omega_0 = 5\; \text{rad/s},\; \delta = 8\; \text{s}^{-1}) | Two exponentials: one decays quickly, the other slowly. The motion is slower than the critical case even though (\delta) is larger. |
| If you plot the envelope (e^{-\delta t}) together with the actual overdamped solution, you will see that the solution never follows the envelope; instead the slower exponential (with rate ( | \lambda_{\text{slow}} | = \delta-\sqrt{\delta^{2}-\omega_0^{2}})) dictates the long‑time behavior. Because ( | \lambda_{\text{slow}} | < \delta), a larger (\delta) can reduce the overall decay rate, giving the “less steep” appearance. |
5. Answering the three specific doubts
5.1 Why the given formula only works for under‑damping
The expression
[ y(t)=y_0 e^{-\delta t}\cos(\omega t) ]
was derived after solving (1) under the condition (\delta<\omega_0). In that regime the square‑root in (2) is imaginary, leading to a sinusoidal factor with angular frequency
[ \omega = \omega_D = \sqrt{\omega_0^{2}-\delta^{2}} . ]
If (\delta\ge\omega_0) the square‑root becomes real, the cosine term would become a hyperbolic cosine or a sum of exponentials, not a simple cosine. Hence the formula is not valid for critical or over‑damping.
5.2 Correct condition for each regime
| Regime | Correct inequality (using the undamped frequency (\omega_0)) |
|---|---|
| Underdamped | (\boxed{\delta < \omega_0}) |
| Critical | (\boxed{\delta = \omega_0}) |
| Overdamped | (\boxed{\delta > \omega_0}) |
If your textbook or teacher writes “δ = ω” they are implicitly meaning “δ equals the undamped ω”.
5.3 Why a larger δ can make the motion slower (over‑damped case)
For overdamping the solution is
[ y(t)=A e^{\lambda_1 t}+B e^{\lambda_2 t},\qquad \lambda_{1,2}= -\delta \pm \sqrt{\delta^{2}-\omega_0^{2}} . ]
Both (\lambda_{1,2}) are negative. The slower (less negative) root is
[ \lambda_{\text{slow}} = -\delta + \sqrt{\delta^{2}-\omega_0^{2}} . ]
A quick algebraic check shows
[ |\lambda_{\text{slow}}| = \delta - \sqrt{\delta^{2}-\omega_0^{2}} \;<\; \delta . ]
Thus the dominant exponential decays with a rate smaller than (\delta). As (\delta) grows much larger than (\omega_0),
[ \lambda_{\text{slow}} \approx -\frac{\omega_0^{2}}{2\delta}, ]
so the decay actually becomes slower the larger the damping constant—exactly what you observed in the simulations.
6. Final Summary
- The damped harmonic oscillator obeys (\ddot y + 2\delta\dot y + \omega_0^{2}y = 0).
-
The three regimes are distinguished by the relation between the damping constant (\delta) and the undamped natural frequency (\omega_0):
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Underdamped (\delta < \omega_0) → oscillatory with envelope (e^{-\delta t}) and damped frequency (\omega_D = \sqrt{\omega_0^{2}-\delta^{2}}).
-
Critically damped (\delta = \omega_0) → (y(t)=(A+Bt)e^{-\delta t}); fastest non‑oscillatory return.
-
Overdamped (\delta > \omega_0) → (y(t)=A e^{\lambda_1 t}+B e^{\lambda_2 t}) (both terms decay, the slower one has rate ( \lambda_{\text{slow}} <\delta)).
-
-
The formula (y(t)=y_0 e^{-\delta t}\cos(\omega t)) is only the under‑damped solution, where (\omega) is the damped angular frequency (\omega_D).
- Increasing (\delta) beyond the critical value makes the dominant decay exponent smaller, so the motion appears “less steep” even though the physical damping coefficient is larger.
7. Common Mistakes
| Mistake | Why it’s wrong | How to avoid it | ||
|---|---|---|---|---|
| Using the same symbol (\omega) for both the undamped and the damped frequency. | The under‑damped solution uses (\omega_D=\sqrt{\omega_0^{2}-\delta^{2}}). Confusing the two leads to the wrong condition for critical damping. | Keep the notation: (\omega_0) = natural (undamped) frequency; (\omega_D) = damped frequency. | ||
| Applying the under‑damped formula (e^{-\delta t}\cos(\omega t)) to the overdamped case. | For (\delta\ge\omega_0) the cosine becomes a hyperbolic cosine or a sum of exponentials; the formula no longer solves the ODE. | First check the inequality (\delta) vs. (\omega_0). Use the appropriate general solution from the roots (2). | ||
| Thinking that a larger (\delta) always makes the system return faster. | In the overdamped regime the dominant decay rate is ( | \lambda_{\text{slow}} | = \delta -\sqrt{\delta^{2}-\omega_0^{2}}), which decreases as (\delta) increases beyond (\omega_0). | Remember that “critical damping” is the fastest non‑oscillatory return; any increase beyond that slows the approach. |
| Ignoring the constant prefactors (A) and (B). | The shape of the response (whether it crosses zero, how quickly it decays) depends on the initial conditions encoded in (A) and (B). | Write the full solution with two arbitrary constants, then apply the given initial displacement and velocity to determine them. | ||
| Misreading “critical damping = δ = ω” as a universal rule. | The teacher’s statement uses (\omega) to mean the undamped frequency (\omega_0). | When you see “δ = ω”, ask whether the instructor is referring to (\omega_0) |
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