Consider the boundary-value problem introduced in the construction of the mathematical model for the shape of a rotating string: T d2y dx2 +

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.)

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Ngọc Hoa 3 years 2021-09-03T07:33:31+00:00 1 Answers 4 views 0

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  1. dangkhoa
    0
    2021-09-03T07:34:45+00:00
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    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}

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