Question

When the magnitude of tension is not constant, then a model for the deflection curve or shape$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}$assumed by a rotating string is given by

localid="1663939956362" $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left[T\left(x\right)\frac{dy}{dx}\right]\mathbf{+}\mathit{\rho }{\mathit{\omega }}^{\mathbf{2}}\mathit{y}\mathbf{=}\mathbf{0}$

Suppose that$\mathbf{1}\mathbf{<}\mathit{x}\mathbf{<}\mathit{e}$and that$\mathbf{T}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}$.

(a). If $\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$and $\mathit{\rho }{\mathit{\omega }}^{\mathbf{2}}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{25}$,

Show that the critical speeds of angular rotation are${\mathit{\omega }}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\sqrt{\frac{\left(4n{}^2{\pi }^2+1\right)}{\mathbf{\rho }}}$and the corresponding deflections are

${\mathit{y}}_{\mathbf{n}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{2}}{\mathit{x}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{n}\mathit{\pi }\mathbf{I}\mathit{n}\mathit{x}\mathbf{\right)}\mathbf{,}$$\mathit{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$

(b). use a graphing utility to graph the deflection curves on the interval $\left[1,e\right]$for$\mathit{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$. Choose${\mathit{c}}_{\mathbf{2}}\mathbf{=}\mathbf{1}$.

Short answer

Finally we get the solutions as

${\omega }_n=\frac{1}{2}\sqrt{\frac{\left(4n{}^2{\pi }^2+1\right)}{\rho }},n=1,2,...$

$y_n\left(x\right)=c_2{x}^{-\frac{1}{2}}\sin \left(n\pi \mathrm{I}nx\right)$

Where$n=1,2,3,...$

Step-by-step solution

To Cauchy-Euler equation

(a)When

$\mathrm{T}\left(x\right)=x^2$

The given differential equation is the Cauchy-Euler equation

$x^2{y}^{\text{'}\text{'}}+2x{y}^{\text{'}}+\rho {\omega }^{2}y=0.$

The solutions of the auxiliary equation

$m(m-1)+2m+\rho {\omega }^2=m^2+m+\rho {\omega }^2=0$

We get

$m_1=-\frac{1}{2}\pm \frac{1}{2}\sqrt{4\rho {\omega }^2-1i}$

Since$\rho {\omega }^2>0.25$

Then ,We get

$y=c_1{x}^{-\frac{1}{2}}\cos \left(\lambda \mathrm{I}nx\right)+c_2{x}^{-\frac{1}{2}}\sin \left(\lambda \mathrm{I}nx\right)$

Where$\lambda =\frac{1}{2}\sqrt{4\rho {\omega }^2-1}$

Applying$y\left(1\right)=0$

That gives $c_1=0$and consequently

$y=c_2{x}^{-\frac{1}{2}}\sin \left(\lambda \mathrm{I}nx\right)$

The condition $y\left(e\right)=0$ Requires$c_2{e}^{-\frac{1}{2}}\sin \lambda =0$ .

Then we get the nontrivial solution when

${\lambda }_n=n\pi ,n=1,2,3,...$

${\lambda }_n=\frac{1}{2}\sqrt{4\rho {\omega }_n^2-1}=n\pi$

And we obtain

${\omega }_n=\frac{1}{2}\sqrt{\frac{\left(4n{}^2{\pi }^2+1\right)}{\rho }},n=1,2,...$

The corresponding solutions are

$y_n\left(x\right)=c_2{x}^{-\frac{1}{2}}\sin \left(n\pi \mathrm{I}nx\right)$

Graph the solution

(b) Graphing the solutions

For $n=1$

For n = 2

For n = 3

Final answer

Finally we get the solutions as

${\omega }_n=\frac{1}{2}\sqrt{\frac{\left(4n{}^2{\pi }^2+1\right)}{\rho }},n=1,2,...$

$y_n\left(x\right)=c_2{x}^{-\frac{1}{2}}\sin \left(n\pi \mathrm{I}nx\right)$

Where$n=1,2,3,...$