Question

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.)

Answer 1

Answer:

$y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x$

$y_n(x) = C_n \sin \frac{n \pi x}{L}$

Explanation:

The given differential equation is

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

where T and ρ  are constants

The given rewrite as

$\frac{d^2y}{dx^2} + \frac{\rho w^2}{T} y=0$

auxiliary equation is

$m^2+ \frac{\rho w^2}{T} =0\\\\m= \pm\sqrt{\frac{\rho}{T} } wi$

Solution of this de is

$y(x)=C_1 \cos \sqrt{\frac{\rho}{t} } wx + C_2 \sin \sqrt{\frac{\rho}{T} } wx$

y(0)=0 ⇒ C₁ = 0

$y(x) = C_2 \sin \sqrt{\frac{\rho}{T} } wx$

y(L) = 0 ⇒

$C_2 \sin \sqrt{\frac{\rho}{T} } wL=0$

we need non zero solution

⇒ C₂ ≠ 0 and

$\sin \sqrt{\frac{\rho}{T} } wL=0$

$\sin \sqrt{\frac{\rho}{T} } wL=0 \rightarrow \sqrt{\frac{\rho}{T} } wL=n \pi$

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

solution corresponding these $w_n$ values

$y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x$

$y_n(x) = C_n \sin \frac{n \pi x}{L}$