Vibrations of an Elastic Body in a Gravitational Field

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Consider an elastic rope with elastic modulus \(Y\), unstretched length \(l\) (gravity-free state), cross-sectional area \(S_0\), linear mass density \(\lambda\), and Poisson’s ratio \(\frac{1}{2}\) (i.e., the medium is incompressible). One end is fixed at point O, while the other end is attached to a mass \(M\). The entire system hangs vertically under gravitational acceleration \(g\). Only longitudinal vibrations of the elastic rope are considered.

Define the unstretched coordinate \(x\) as the distance from O to a point on the rope in its free state. The elongation coordinate \(\xi{(x)}\) is the displacement (positive downward) at the point where the unstretched coordinate is \(x\).

Tension Distribution \(F(x)\) on the Rope #

The elastic rope is incompressible, hence

$$ S\cdot(\mathrm{d}x+\mathrm{d}\xi)=S_0\cdot\mathrm{d}x \tag{1} $$

From Hooke’s law

$$ \frac{F}{S}=Y\frac{\partial \xi}{\partial x} \tag{2} $$

Therefore

$$ F(x)=S_0 Y\frac{\frac{\partial \xi}{\partial x}}{1+\frac{\partial \xi}{\partial x}} \tag{3} $$

Wave Equation #

For an infinitesimal element \(x\,{\backsim}\,x+dx\), we have

$$ \lambda dx\cdot\frac{\partial^2\xi}{\partial t^2}=\lambda dx\cdot g+F(x+dx)-F(x) \tag{4} $$

Substituting (3) yields

$$ \lambda\frac{\partial^2\xi}{\partial t^2}=\lambda g+\frac{S_0Y}{(1+\frac{\partial\xi}{\partial x})^2}\frac{\partial^2\xi}{\partial x^2} \tag{5} $$

Boundary conditions

$$ \begin{cases} \xi|_{x=0}=0\\ M\cdot\frac{\partial^2\xi}{\partial t^2}|_{x=l}=Mg-S_0Y\frac{(\frac{d\xi}{dx})|_{x=l}}{1+(\frac{d\xi}{dx})|_{x=l}} \end{cases} \tag{6} $$

Steady-State Solution #

The steady-state solution without vibration is \(\xi_0(x)\), where \(\frac{\partial^2\xi _0}{\partial t^2}=0\). Thus, (5) simplifies to

$$ \lambda g+\frac{S_0Y}{(1+\frac{d\xi _0}{dx})^2}\frac{d^2\xi _0}{d x^2} = 0 \tag{7} $$

Integrating gives

$$ \frac{d\xi_0 }{dx}=\frac{Mg+\lambda (l-x)g }{S_0Y-Mg-\lambda( l-x)g } \tag{8} $$

Hence

$$ \xi_0 \left ( x \right ) =\frac{S_0Y}{\lambda g} \ln\left ( \frac{S_0Y-Mg-\lambda lg+\lambda xg}{S_0Y-Mg-\lambda lg} \right ) -x \tag{9} $$

Small-Amplitude Vibration Approximation #

For small-amplitude vibrations, define

$$ \xi(x,t)=\xi_0(x)+\delta(x,t) \quad , |\delta|\ll|\xi_0| \tag{10} $$

Substituting into (5) and retaining first-order terms gives

$$ \lambda\frac{\partial^2\xi}{\partial t^2}=\lambda g+\frac{S_0Y}{(1+\frac{d\xi_0}{dx})^2}\left[\frac{d^2\xi_0}{dx^2}+\frac{\partial^2\delta}{\partial x^2}-2\frac{d^2\xi_0}{dx^2}\frac{\frac{\partial\delta}{\partial x}}{1+\frac{d\xi_0}{dx}}\right] \tag{11} $$

Substituting (9) and simplifying yields the first-order approximated equation for small vibrations

$$ \lambda\frac{\partial^2\xi}{\partial t^2}=\frac{[S_0Y-Mg-\lambda g(l-x)]^2}{S_0Y}\frac{\partial^2\delta}{\partial x^2}+2\lambda g\frac{S_0Y-Mg-\lambda g(l-x)}{S_0Y}\frac{\partial\delta}{\partial x} \quad \text{(12)} $$

Separate variables by letting \(\delta(x,t) =A(x)B(t)\)

$$ \frac{1}{B} \lambda \frac{d^2B}{dt^2} =\left \{ \frac{\left [ S_0Y-Mg-\lambda \left ( l-x \right )g \right ]^2}{S_0Y} \frac{d^2A}{dx^2} +2\lambda g\frac{S_0Y-Mg-\lambda \left ( l-x \right )g }{S_0Y} \frac{dA}{dx} \right \} \frac{1}{A} \quad \text{(13)} $$

The left side depends only on \(t\), the right side only on \(x\). Thus, both sides equal a constant, denoted \(-\lambda \omega^2\). The left side gives

$$ \frac{1}{B} \lambda \frac{d^2B}{dt^2}=-\lambda \omega^2 \tag{14} $$

$$ B\left ( t\right )=B_o\cos \left ( \omega t+\varphi \right ) \tag{15} $$

The right side gives

$$ \frac{\left [ S_0Y-Mg-\lambda \left ( l-x \right )g \right ]^2 }{S_0Y} \frac{d^2A}{dx^2} +2\lambda g\frac{S_0Y-Mg-\lambda \left ( l-x \right )g}{S_0Y}\frac{dA}{dx}+\lambda \omega^2A=0 \quad \text{(16)} $$

This is an Euler equation. Let

$$ t=\ln\left ( S_0Y-Mg-\lambda lg+\lambda gx \right ) \tag{17} $$

Substituting back into (16) gives

$$ \frac{d^2A}{dt^2}+\frac{dA}{dt}+\frac{\omega^2S_0Y }{\lambda g^2}A=0 \tag{18} $$

The characteristic equation is

$$ \lambda ^2+\lambda +\frac{\omega^2S_0Y }{\lambda g^2}=0 \tag{19} $$

$$ \lambda_{1,2}=\frac{-1\pm \sqrt{1-4\frac{\omega^2S_0Y}{\lambda g^2} } }{2} \tag{20} $$

Thus

$$ A=A_1e^{\lambda_1t}+A_2e^{\lambda_2t} \tag{21} $$

$$ \delta =\left [ A_1\left ( S_0Y-Mg-\lambda lg+\lambda xg \right )^{\lambda_1}+A_2\left ( S_0Y-Mg-\lambda lg+\lambda xg \right ) ^{\lambda_2} \right ]\cos \left ( \omega t+\varphi \right ) \quad \text{(22)} $$

(The constant \(B_0\) is absorbed into \(A_1,A_2\).)

Applying the first-order approximation to boundary conditions (6)

$$ \begin{cases} \delta|_{x=0}=0\\ M\frac{\partial ^2\delta }{\partial t^2}|_{x=l}=Mg-S_0Y\frac{\frac{d\xi_0}{dx}}{1+\frac{d\xi_0}{dx}}|_{x=l}-S_0Y\frac{\frac{\partial \delta}{\partial x}}{1+\frac{d\xi_0}{dx}}|_{x=l} \end{cases} \tag{23} $$

Substituting (9) to eliminate \(\xi_0\)

$$ \begin{cases} \delta|_{x=0}=0\\ M\frac{\partial ^2\delta }{\partial t^2}|_{x=l}=-\frac{(S_0Y-Mg)^2}{S_0Y}\frac{\partial\delta}{\partial x}|_{x=l} \end{cases} \tag{24} $$

Substituting (22) to eliminate \(\delta\)

$$ {\begin{cases} A_1\left ( S_0Y-Mg-\lambda lg \right ) ^{\lambda _1-\lambda _2}+A_2=0\\ w^2M\left [ A_1\left ( S_0Y-Mg \right )^{\lambda _1-\lambda _2} +A_2 \right ]=\frac{S_0Y-Mg}{S_0Y}\lambda g [ A_1\lambda _1\left ( S_0Y-Mg )^{\lambda _1-\lambda _2}+A_2\lambda _2 \right ] \end{cases}} \quad \text{(25)} $$

This is a homogeneous linear system in \(A_1,\ A_2\). For non-trivial solutions, the determinant must be zero, yielding the equation for \(\omega\)

$$ \omega^2M\left [ \left ( \frac{S_0Y-Mg}{S_0Y-Mg-\lambda lg} \right )^{\lambda _1-\lambda _2} -1 \right ] =\frac{S_0Y-Mg}{S_0Y}\lambda g \left[ \lambda _1\left(\frac{S_0Y-Mg}{S_0Y-Mg-\lambda lg}\right)^{\lambda _1-\lambda _2} -\lambda _2 \right ] \quad \text{(26)} $$

If

$$ 4\frac{\omega^2S_0Y}{\lambda g^2} > 1 \tag{27} $$

then \(\lambda_1,\lambda_2\) are complex. Let

$$ \lambda _{1,2}=\alpha \pm \beta i \tag{28} $$

where

$$ \alpha =-\frac{1}{2} \tag{29} $$

$$ \beta =\sqrt{\frac{\omega^2S_0Y}{\lambda g^2}-\frac{1}{4}} \tag{30} $$

Substituting into (26) gives

$$ \begin{align} \omega^2M =& \frac{S_0Y-Mg}{S_0Y}\lambda g\left \{ \alpha +\beta \cot \left [ \beta \ln\left ( \frac{S_0Y-Mg}{S_0Y-Mg-\lambda lg} \right ) \right ] \right \} \notag \\ =& \frac{S_0Y-Mg}{S_0Y}\lambda g\left \{ -\frac{1}{2} +\sqrt{\frac{\omega^2S_0Y}{\lambda g^2}-\frac{1}{4}}\ \cot \left [ \sqrt{\frac{\omega^2S_0Y}{\lambda g^2}-\frac{1}{4}} \ \ln\left ( \frac{S_0Y-Mg}{S_0Y-Mg-\lambda lg} \right ) \right ] \right \} \notag \quad \text{(31)} \end{align} $$

This is the equation satisfied by \(\omega\).