Question

Consider a driven undammed spring/mass system described by the initial-value problem $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathit{x}\mathbf{=}{\mathit{F}}_{\mathbf{0}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{n}}\mathit{\gamma }\mathit{t}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$(a) For $\mathit{n}\mathbf{=}\mathbf{2}$, discuss why there is a single frequencyrole="math" localid="1664263381179" $\raisebox{1ex}{${\mathbf{\gamma }}_{\mathbf{1}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}\mathbf{\pi }$}\right.$at which the system is in pure resonance. (b) For n = 3$\mathit{n}\mathbf{=}\mathbf{3}$, discuss why there are two frequencies $\raisebox{1ex}{${\mathbf{\gamma }}_{\mathbf{1}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}\mathbf{\pi }$}\right.$ and$\raisebox{1ex}{${\mathbf{\gamma }}_{\mathbf{2}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}\mathbf{\pi }$}\right.$at which the system is in pure resonance. (c) Suppose$\mathit{\omega }\mathbf{=}\mathbf{1}$and ${\mathit{F}}_{\mathbf{0}}\mathbf{=}\mathbf{1}$. Use a numerical solver to obtain the graph of the solution of the initial-value problem for$\mathit{n}\mathbf{=}\mathbf{2}$androle="math" localid="1664263250146" $\mathit{\gamma }\mathbf{=}{\mathit{\gamma }}_{\mathbf{1}}$in part (a). Obtain the graph of the solution of the initial-value problem for$\mathit{n}\mathbf{=}\mathbf{3}$corresponding, $\mathit{n}$turn, to $\mathit{\gamma }\mathbf{=}{\mathit{\gamma }}_{\mathbf{1}}$and$\mathit{\gamma }\mathbf{=}{\mathit{\gamma }}_{\mathbf{2}}$in part (b).

Short answer

(a) For $n=2$, discuss why there is a single frequency $\raisebox{1ex}{${\gamma }_1$}\!\left/ \!\raisebox{-1ex}{$2\pi $}\right.$at which the system is in pure resonance.

(b) For $n=3$, discuss why there are two frequencies $\raisebox{1ex}{${\gamma }_1$}\!\left/ \!\raisebox{-1ex}{$2\pi $}\right.$and $\raisebox{1ex}{${\gamma }_2$}\!\left/ \!\raisebox{-1ex}{$2\pi $}\right.$at which the system is in pure resonance.

(c) The required graph is in the answer.

Step-by-step solution

Definition of Spring / Mass Systems:

Suppose you take into consideration an external force$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$acting on a vibrating mass on a spring. For example,$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$could represent a driving force causing an oscillatory vertical motion of the support of the spring. The inclusion of$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$in the formulation of Newton’s second law gives the differential equation of driven of forced motion:

$\mathit{m}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}\mathbf{-}\mathit{\beta }\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

(a) An undammed system, which consists of mass and spring:

For $n=2$there is a single frequency $\raisebox{1ex}{${\gamma }_1$}\!\left/ \!\raisebox{-1ex}{$2\pi $}\right.$at which the system is in pure resonance.

$\begin{array}{rcl}\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x& =& F_0{\sin }^n\gamma t\\ & =& F_0{\sin }^2\gamma t\\ & =& F_0\frac{1}{2}\times 2{\sin }^2\gamma t\\ & =& \frac{F}{2}[1-\cos 2\gamma t]\end{array}$

For $n=2$,${\sin }^2\gamma t=\frac{1}{2}(1-\cos 2\gamma t)$

The system is in pure resonance when,

$$\begin{gathered} \frac{2{\gamma }_1}{2\pi }=\frac{\omega }{2\pi } \\ {\gamma }_1=\frac{\omega }{2} \end{gathered}$$

(b) For n=3, there are two frequencies γ12π and γ22π  at which the system is in pure resonance:

For $n=3$:

$\begin{array}{rcl}{\sin }^2\gamma t& =& \sin \gamma t{\sin }^2\gamma t\\ & =& \sin \gamma t\times \frac{1}{2}(1-\cos \gamma t)\\ & =& \frac{1}{2}\left[\sin \gamma t(\sin \gamma t-\cos 2\gamma t)\right]\end{array}$

Now,

$\sin \gamma t\cos 2\gamma t=\frac{1}{2}\left[\sin 3\gamma t-\sin \gamma t\right]$

And as you know,

$\sin (A+B)+\sin (A-B)=2\sin A\cos B$

Therefore,

$\begin{array}{rcl}{\sin }^2\gamma t& =& \frac{1}{2}\left[\sin \gamma t-\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin 3\gamma t+\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin \gamma t\right]\\ & =& \frac{3}{4}\sin \gamma t-\frac{1}{4}\sin 3\gamma t\end{array}$

Thus,

$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=\frac{3}{4}\sin \gamma t-\frac{1}{4}\sin 3\gamma t$

The frequency of free vibration is $\frac{\omega }{2\pi }$.

Thus, when

$$\begin{gathered} \frac{{\gamma }_1}{2\pi }=\frac{\omega }{2\pi } \\ {\gamma }_1=\omega \end{gathered}$$

And when

$$\begin{gathered} \frac{3{\gamma }_2}{2\pi }=\frac{\omega }{2\pi } \\ {\gamma }_2=\frac{\omega }{3} \end{gathered}$$

The system will be in pure resonance.

(c) The graph:

For$n=2$:

For$n=3$: