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Consider the boundary-value problem introduced in the construction of the mathematical model for the shape of a rotating string: T d2y dx2 + rhoω2y = 0, y(0) = 0, y(L) = 0. For constants T and rho, define the critical speeds of angular rotation ωn as the values of ω for which the boundary-value problem has nontrivial solutions. Find the critical speeds ωn and the corresponding deflections yn(x). (Give your answers in terms of n, making sure that each value of n corresponds to a unique critical speed.)

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shallomisaiah19

Answer:

[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]

[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]

Explanation:

The given differential equation is

[tex]T\frac{d^2y}{dx^2} + \rho w ^2y=0[/tex] and y(0) = 0, y(L) =0

where T and ρ  are constants

The given rewrite as

[tex]\frac{d^2y}{dx^2} + \frac{\rho w^2}{T} y=0[/tex]

auxiliary equation is

[tex]m^2+ \frac{\rho w^2}{T} =0\\\\m= \pm\sqrt{\frac{\rho}{T} } wi[/tex]

Solution of this de is

[tex]y(x)=C_1 \cos \sqrt{\frac{\rho}{t} } wx + C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]

y(0)=0 ⇒ C₁ = 0

[tex]y(x) = C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]

y(L) = 0 ⇒

[tex]C_2 \sin \sqrt{\frac{\rho}{T} } wL=0[/tex]

we need non zero solution

⇒ C₂ ≠ 0 and

[tex]\sin \sqrt{\frac{\rho}{T} } wL=0[/tex]

[tex]\sin \sqrt{\frac{\rho}{T} } wL=0 \rightarrow \sqrt{\frac{\rho}{T} } wL=n \pi[/tex]

[tex]w_n = \sqrt{\frac{T}{\rho} } \frac{n \pi}{L}[/tex]

solution corresponding these [tex]w_n[/tex] values

[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]

[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]

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