Question

Consider the boundary – value problem introduced in the construction of the mathematical model for the shape of a rotating string

$\mathbf{T}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\mathit{\rho }{\mathit{\omega }}^{\mathbf{2}}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathit{L}\mathbf{\right)}\mathbf{=}\mathbf{0}$

For constant $c_2$and , define the Critical speeds of angular rotation ${\mathit{\omega }}_{\mathbf{n}}$as the values of $\mathit{\omega }$for which the boundary – value problem has nontrivial solutions. Find the critical speeds ${\mathit{\omega }}_{\mathbf{n}}$and the corresponding deflections${\mathit{y}}_{\mathbf{n}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$.

Short answer

Therefore, the answer is${\omega }_n=\frac{n\pi }{L}\sqrt{\frac{\rho }{T}},n=1,2,\cdots$

and

$y_n\left(x\right)=c_2\sin \frac{n\pi x}{L}$

where$c_2$ is a nonzero constat.

Step-by-step solution

Given Information.

The given solution is:

$\mathrm{T}\frac{d^{2}y}{d{x}^2}+\rho {\omega }^{2}y=0,y\left(0\right)=0,y\left(L\right)=0$

Find the critical speed

The general solution is

$y\left(x\right)=c_1\cos \sqrt{\frac{\rho }{\mathrm{T}}}\omega x+c_2\sin \sqrt{\frac{\rho }{\mathrm{T}}}\omega x.$

From the initial conditions

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

We get

$c_1=0$

Setting $y\left(L\right)=0$we get

$c_2\sin \sqrt{\frac{\rho }{\mathrm{T}}}\omega L$

Putting$c_2\ne 0$ to get a nontrivial solution and

We get

$\sqrt{\frac{\rho }{\mathrm{T}}}\omega L=n\pi$

And so

${\omega }_n=\frac{n\pi }{L}\sqrt{\frac{\rho }{\mathrm{T}}},n=1,2,.....$

The deflections are

$y_n\left(x\right)=c_2\sin \frac{n\pi x}{L}.$

Where $c_2$is a nonzero constant.

Find the corresponding deflection

Finally we get the solution as

${\omega }_n=\frac{n\pi }{L}\sqrt{\frac{\rho }{\mathrm{T}}},n=1,2,....$

And

$y_n\left(x\right)=c_2\sin \frac{n\pi x}{L}.$

Where is a nonzero constant.