Question

(a) Show that the solution of initial-value problem $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathit{x}\mathbf{=}{\mathit{F}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\mathit{\gamma }\mathit{t}\mathbf{,}\mathbf{\text{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\text{x'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$is $\mathit{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{0}}}{{\mathbf{\omega }}^{\mathbf{2}}\mathbf{-}{\mathbf{\gamma }}^{\mathbf{2}}}\mathbf{(}\mathbf{cos\gamma t}\mathbf{-}\mathbf{cos\omega t}\mathbf{)}$.

(b) Evaluate role="math" localid="1664189755201" $\underset{\mathbf{\gamma }\mathbf{\to }\mathbf{\omega }}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{{\mathbf{F}}_{\mathbf{0}}}{{\mathbf{\omega }}^{\mathbf{2}}\mathbf{-}{\mathbf{\gamma }}^{\mathbf{2}}}\mathbf{(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\gamma }\mathbf{t}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\omega }\mathbf{t}\mathbf{)}$.

Short answer

(a)$x\left(t\right)=\frac{F_0}{{\omega }^2-{\gamma }^2}(\cos \gamma t-\cos \omega t)$

(b) Evaluation of $\underset{\gamma \to \omega }{\lim }\frac{F_0}{{\omega }^2-{\gamma }^2}(\cos \gamma t-\cos \omega t)$is $\frac{F_0}{2\omega }t\sin \omega t$.

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) Differential equation:

The solution of the given differential equation is given by,

$x\left(t\right)=x_c\left(t\right)+x_p\left(t\right)$

Where the complimentary function $x_c\left(t\right)$is the solution to the homogenous differential equation that is,

$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$

And the particular function $x_p\left(t\right)$is the solution to be,

$x_p^{\text{'}\text{'}}+{\omega }^2{x}_p=F_0\cos \gamma t$

Solving for $x_c\left(t\right)$, you have the auxiliary equation as,

$m^2+{\omega }^2=0$

This gives,

$m_{1,2}=0\pm \omega i$

Solve Complementary and particular function:

Thus, the general solution to (1) is,

$x_c\left(t\right)=c_1\cos \omega t+c_2\sin \omega t$

Using undetermined coefficients, you have $x_p$and its derivatives to be,

$$\begin{gathered} x_p=A\cos \gamma t+B\sin \gamma t \\ {x}_p^{\text{'}}=-A\gamma \sin \gamma t+B\gamma \cos \gamma t \\ {x^{\text{'}\text{'}}}_p=-A{\gamma }^2\cos \gamma t-B{\gamma }^2\sin \gamma t \end{gathered}$$

Following equation (2) in Step 3, you have

$\begin{array}{rcl}{F}_0\cos \gamma t& =& -A{\gamma }^2\cos \gamma t-B{\gamma }^2\sin \gamma t+{\omega }^2(A\cos \gamma t+B\sin \gamma t)\\ & =& ({\omega }^2-{\gamma }^2)A\cos \gamma t+({\omega }^2-{\gamma }^2)B\sin \gamma t\end{array}$

Equating the coefficients of the resulting equation in Step 3 gives,

$$\begin{gathered} ({\omega }^2-{\gamma }^2)A=F_0 \\ A=\frac{F_0}{{\omega }^2-{\gamma }^2} \end{gathered}$$

And

$$\begin{gathered} ({\omega }^2-{\gamma }^2)B=0 \\ B=0 \end{gathered}$$

Thus, the particular solution to (2) is,

$\begin{array}{rcl}{x}_p\left(t\right)& =& A\cos \gamma t+B\sin \gamma t\\ & =& \frac{F_0}{{\omega }^2-{\gamma }^2}\cos \gamma t\end{array}$

Now that you have solved for $x_c\left(t\right)$and $x_p\left(t\right)$, it follows that,

$x\left(t\right)=c_1\cos \omega t+c_2\sin \omega t+\frac{F_0}{{\omega }^2-{\gamma }^2}\cos \gamma t$

Substitute the initial condition:

However, there are still unknowns. You need to utilize the initial conditions to be able to find$c_1$and $c_2$.

You are given with the initial condition$x\left(0\right)=0$.

Solving for , you get

$\begin{array}{rcl}0& =& c_1\cos 0°+c_2\sin 0°+\frac{F_0}{{\omega }^2-{\gamma }^2}\cos 0°\\ & =& c_1\left(1\right)+c_2\left(0\right)+\frac{F_0}{{\omega }^2-{\gamma }^2}\end{array}$

$c_1=-\frac{F_0}{{\omega }^2-{\gamma }^2}$

Using the first derivative ofand the other initial condition$x^{\text{'}}\left(0\right)=0$to find$c_2$gives,

$$\begin{gathered} x\left(t\right)=c_1\cos \omega t+c_2\sin \omega t+\frac{F_0}{{\omega }^2-{\gamma }^2}\cos \gamma t \\ {x}^{\text{'}}\left(t\right)=-\omega c_1\sin \omega t+\omega c_2\cos \omega t-\frac{F_0}{{\omega }^2-{\gamma }^2}\gamma \sin \gamma t \\ 0=-\omega \left(\frac{F_0}{{\omega }^2-{\gamma }^2}\right)\sin 0°+\omega c_2\cos 0°-\frac{F_0}{{\omega }^2-{\gamma }^2}\gamma \sin 0° \\ {c}_2=0 \end{gathered}$$

With the values of $c_1$and $c_2$obtained in Steps 6 and 7, we have the solution to the initial-value problem to be,

$\begin{array}{rcl}x\left(t\right)& =& c_1\cos \omega t+c_2\sin \omega t+\frac{F_0}{{\omega }^2-{\gamma }^2}\cos \gamma t\\ & =& -\frac{F_0}{{\omega }^2-{\gamma }^2}\cos \omega t+\frac{F_0}{{\omega }^2-{\gamma }^2}\cos \gamma t\\ & =& \frac{F_0}{{\omega }^2-{\gamma }^2}(\cos \gamma t-\cos \omega t)\end{array}$

(b) Using L'Hôpital's rule:

Taking the limit of the solution directly as$\gamma \to \omega$leads to an indeterminate form,

$\begin{array}{rcl}\underset{\gamma \to \omega }{\lim }\frac{F_0}{{\omega }^2-{\gamma }^2}(\cos \gamma t-\cos \omega t)& =& \frac{F_0}{{\omega }^2-{\omega }^2}(\cos \omega t-\cos \omega t)\\ & =& 0\end{array}$

Using L'Hôpital's rule, you find

$\begin{array}{rcl}\underset{\gamma \to \omega }{\lim }\frac{F_0}{{\omega }^2-{\gamma }^2}(\cos \gamma t-\cos \omega t)& =& F_0\underset{\gamma \to \omega }{\lim }\frac{{\displaystyle \frac{d}{d{\displaystyle \gamma }}}(\cos \gamma t-\cos \omega t)}{{\displaystyle \frac{d}{d{\displaystyle \gamma }}}({\omega }^2-{\gamma }^2)}\\ & =& F_0\underset{\gamma \to \omega }{\lim }\frac{-t\sin \gamma t-0}{0-2\gamma }\\ & =& \frac{F_0}{2\omega }t\sin \omega t\end{array}$