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A driven oscillator on explicit data: the steady response, its amplitude maximum and a worked transient at $m = k = 1$ , $f_{0} = 1$ , $c = 0.2$ , $Ω = 1$

Physical scope

A worked check of the driven oscillator at the declared data m=k=1, f_0=1 N, c=0.2 kg/s, Omega=1 rad/s: the steady response 5 sin t and its verification by substitution, the amplitude 5 m at the drive frequency against the maximum value of the amplitude function and its location, the phase lag, the transient motion from rest and its exponential approach to the steady response, and the comparison with the undamped cases. The item asserts no measured value and no claim about a physical apparatus.

Example

Declare the sinusoidally driven oscillator model of The sinusoidally driven oscillator model: a prescribed external force $f_{0} \cos (Ω t)$ added to the damped oscillator with $m = 1 k g, k = 1 N / m, f_{0} = 1 N, c = \frac{1}{5} k g / s, Ω = 1 s^{- 1},$ so that the physical natural frequency is $\sqrt{k / m} = 1 s^{- 1}$ . In the displayed calculations, $t$ denotes the numerical SI time $t_{p h y s} / (1 s)$ , and position formulas give numerical values in metres. Unless a physical unit is explicitly attached, parameters in those calculations also mean SI numerical values: $m, k, c, f_{0}$ are normalized by $1 k g$ , $1 N / m$ , $1 k g / s$ and $1 N$ , respectively, and $Ω, ω_{0}, ω_{d}, γ$ mean physical rates divided by $1 s^{- 1}$ . Thus the numerical drive is $f (t) = \cos t$ , and derivatives of numerical position with respect to $t$ give numerical velocity in metres per second. The claims checked below are:

the steady response is $x_{p} (t) = 5 \sin t$ metres, a motion on $R$ with amplitude $A (1) = 5$ m and phase lag $δ = π / 2$ ;
the amplitude function $A (Ω) = 1 / \sqrt{(1 - Ω^{2})^{2} + Ω^{2} / 25}$ attains its maximum at $Ω_{*} = \sqrt{0.98} = \sqrt{49 / 50}$ with the value $A (Ω_{*}) = 10 / \sqrt{3.96}$ , larger than $A (1) = 5$ ;
the motion with datum $(0, 0, 0)$ is $x (t) = 5 \sin t - \frac{5}{ω_{d}} e^{- t / 10} \sin (ω_{d} t), ω_{d} = \sqrt{0.99} = \sqrt{99} / 10,$ which is bounded on $[0, \infty)$ and satisfies $x (t) - x_{p} (t) \to 0$ and $x^{'} (t) - x_{p}^{'} (t) \to 0$ as $t \to + \infty$ ;
at $c = 0$ the model with the same drive at $Ω = 1$ has the unbounded motion $t \sin t / 2$ , while at $Ω = 2$ every motion is bounded with amplitude $1 / 3$ m for the steady response.

Verification

Proof technique: direct; substitute the displayed steady response into the equation, evaluate the amplitude function and its maximum, and solve the initial value problem explicitly.

Given: The driven model with $m = k = 1$ , $f_{0} = 1$ , $c = 1 / 5$ , $Ω = 1$ , the steady response $x_{p} = 5 \sin t$ , and the amplitude function $A$ .

[L1] General theory (The driven oscillator: steady response and amplitude profile, bounded damped and off-resonant motions, and unbounded undamped resonance, Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at $Ω = \sqrt{k / m}$ , The amplitude of the sinusoidally forced damped oscillator: location and value of its maximum, The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data): for this model $D = k - m Ω^{2} + i c Ω$ , $x_{p} = Re (f_{0} e^{i Ω t} / D)$ when $D \neq 0$ , the amplitude is $A (Ω) = ∣ f_{0} ∣ / \sqrt{(k - m Ω^{2})^{2} + c^{2} Ω^{2}}$ , every motion is $x_{p} +$ homogeneous, the transient decays for $c > 0$ , every maximal motion is bounded on $[0, \infty)$ for $c > 0$ , and for $c = 0$ the exceptional frequency is $Ω = ω_{0}$ with the unbounded response $f_{0} t \sin (ω_{0} t) / (2 m ω_{0})$ and bounded motions otherwise. For $c^{2} < 4 m k$ , the homogeneous family is $e^{- γ t} (A \cos (ω_{d} t) + B \sin (ω_{d} t))$ , with $γ = c / (2 m)$ and $ω_{d} = \sqrt{k / m - γ^{2}}$ .

[L2] Derivatives and values: $(\sin t)^{'} = \cos t$ , $(\cos t)^{'} = - \sin t$ (The derivatives of sine and cosine are cosine and minus sine), $(e^{μ t})^{'} = μ e^{μ t}$ (The exponential function is smooth and $(\exp)^{'} = \exp$ ), products by the algebra of derivatives and composites by the chain rule; $\sin (π / 2) = 1$ , $\cos (π / 2) = 0$ (Quarter-turn values and shifts by pi/2 and pi); $\sin^{2} + \cos^{2} = 1$ and $∣ \cos ∣ \leq 1$ (Parity and the Pythagorean identity for sine and cosine).

[L3] Dimensions of the underlying physical quantities (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework): $[m] = μ$ , $[k] = μ τ^{- 2}$ , $[c] = μ τ^{- 1}$ , $[f_{0}] = μ ℓ τ^{- 2}$ , $[Ω] = τ^{- 1}$ ; the denominator $D$ has the dimension of $k$ , so $f_{0} / D$ carries length.

1.1givenL1L2L3algebra

Parameters: $ω_{0} = 1$ , $c Ω = 1 / 5$ , $k - m Ω^{2} = 0$ , so $D = i / 5 \neq 0$ , $∣ D ∣ = 1 / 5$ and $A (1) = ∣ f_{0} ∣ / ∣ D ∣ = 5$ m. The general real form gives $x_{p} (t) = \cos t \cdot 0 + \sin t \cdot (1 \cdot (1 / 5)) / (1 / 25) = 5 \sin t$ , and the substitution check is direct: $x_{p}^{''} + c x_{p}^{'} + k x_{p} = - 5 \sin t + \cos t + 5 \sin t = \cos t$ by [L2], so $x_{p}$ is a motion on $R$ . Taking $δ = π / 2$ gives $x_{p} (t) = 5 \cos (t - π / 2)$ with $\cos δ = 0$ and $\sin δ = 1$ : the phase lag is $π / 2$ .

2.1givenstep 1.1L1algebra

Amplitude maximum: $A (Ω) = 1 / \sqrt{(1 - Ω^{2})^{2} + Ω^{2} / 25}$ , so its continuous extension has $A (0) = 1$ and $A (1) = 5$ . By [L1] and $c^{2} = 1 / 25 < 2 m k = 2$ , the maximum occurs at $Ω_{*} = \sqrt{k / m - c^{2} / (2 m^{2})} = \sqrt{1 - 1 / 50} = \sqrt{49 / 50}$ and equals $2 m ∣ f_{0} ∣ / (c \sqrt{4 m k - c^{2}}) = 10 / \sqrt{99 / 25} = 50 / (3 \sqrt{11}) > 5$ , since $3 \sqrt{11} < 10$ because $99 < 100$ ; hence the drive at the natural frequency is below the peak of the amplitude curve.

2.2givenstep 1.1L1L2algebra

Transient from rest. By [L1] every motion equals $x_{p} + y$ with $y$ homogeneous; with numerical rates $γ = c / (2 m) = 1 / 10$ and $ω_{d} = \sqrt{k / m - γ^{2}} = \sqrt{99} / 10$ , the homogeneous motions in numerical time are $y = e^{- t / 10} (A \cos (ω_{d} t) + B \sin (ω_{d} t))$ . The datum $(0, 0, 0)$ forces $A = 0$ and $5 + ω_{d} B = 0$ by [L2], that is $x (t) = 5 \sin t - (5 / ω_{d}) e^{- t / 10} \sin (ω_{d} t)$ , the displayed motion; it is the unique maximal motion with that datum by clause 2 of The driven oscillator: steady response and amplitude profile, bounded damped and off-resonant motions, and unbounded undamped resonance.

3.1step 2.2L1L2

Boundedness and convergence: for $t \geq 0$ , $∣ x (t) ∣ \leq 5 + (5 / ω_{d}) e^{- t / 10} \leq 5 + 50 / \sqrt{99} < 11$ , since $\sqrt{99} > 9$ by [L2], and $x (t) - x_{p} (t) = - (5 / ω_{d}) e^{- t / 10} \sin (ω_{d} t) \to 0$ as $t \to + \infty$ ; differentiating the transient term with [L2] shows that it and its derivative both vanish at infinity, so $x^{'} (t) - x_{p}^{'} (t) \to 0$ ; this is the convergence clause 3 of The driven oscillator: steady response and amplitude profile, bounded damped and off-resonant motions, and unbounded undamped resonance, with the explicit decay rate $1 / 10$ .

4.1givenstep 1.1L1L2algebra∎

Undamped comparison. With $c = 0$ the drive at $Ω = ω_{0} = 1$ is the exceptional case, and clause 5 of The driven oscillator: steady response and amplitude profile, bounded damped and off-resonant motions, and unbounded undamped resonance gives the unbounded motion $t \sin t / 2$ ; with $Ω = 2$ one has $D = 1 - 4 = - 3$ and the steady amplitude $A (2) = 1 / ∣ D ∣ = 1 / 3$ m, with every motion bounded by clause 4 of the same theorem.

Remarks

What the check exhibits. At the drive frequency equal to the natural frequency of the undamped model, the damped model's steady amplitude is $5$ m, and the peak of the amplitude curve is at $Ω_{*} = \sqrt{0.98}$ with value $10 / \sqrt{3.96} \approx 5.025$ m; the transient from rest dies with rate $1 / 10$ s $^{- 1}$ . The dimensionless ratio $A (ω_{0}) / A (0) = 5$ is $ω_{0} / (2 γ)$ in the notation of the source's quality factor. All decimal numbers here are approximations of the exact expressions displayed.
Scope. The data $m = k = f_{0} = 1$ , $c = 1 / 5$ , $Ω = 1$ are declared model data of this check; no measured value is asserted and no apparatus is claimed to be described by the model.

A driven oscillator on explicit data: the steady response, its amplitude maximum and a worked transient at $m = k = 1$ , $f_{0} = 1$ , $c = 0.2$ , $Ω = 1$

Example

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A driven oscillator on explicit data: the steady response, its amplitude maximum and a worked transient at m=k=1m=k=1, f0=1f0​=1, c=0.2c=0.2, Ω=1Ω=1

Example

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A driven oscillator on explicit data: the steady response, its amplitude maximum and a worked transient at $m = k = 1$ , $f_{0} = 1$ , $c = 0.2$ , $Ω = 1$