练习13

一根均质、等截面杆，左端固定，右端自由。一附加质量(视为质点)位于距左端$s$处，$M=\rho AL $。绘制系统前四阶固有频率随$s$变化的曲线。
png

可以列出此杆的振动方程：
$$\frac{\partial^2 u}{\partial t^2}=a^2\frac{\partial ^2 u}{\partial x^2} \qquad x\in(0,L)$$
其中$$a=\sqrt{\frac{E}{\rho}}$$

但由于杆中有集中质量，在$x\in (0,s)$和$x\in (s,L)$上振动的边界条件不同，所以需要分开讨论。

[1]
$$\frac{\partial ^2 u_1}{\partial t^2}=a^2\frac{\partial ^2 u_1}{\partial x^2} \qquad x\in(0,s)$$

[2]
$$\frac{\partial ^2 u_2}{\partial t^2}=a^2\frac{\partial ^2 u_2}{\partial x^2} \qquad x\in(s,L)$$

边界条件:
$$u_1(0,t)=0 \qquad (1)$$

$$EA \left. \frac{\partial u_1}{\partial x} \right | _{x=s^-} + M \left. \frac{\partial ^2 u_1}{\partial t^2} \right | _{x=s^-}=EA\left. \frac{\partial u_2 }{\partial x} \right | _{x=s^+} \qquad (2)$$

$$EA\left. \frac{\partial u_2 }{\partial x} \right | _{x=L}=0 \qquad (3)$$

$$u_1(s^-,t)=u_2(s^+,t) \qquad (4)$$

主振动的一般形式为:
$$u(x,t)=U(x)b \ \sin(\omega t+\phi)$$

那么对于$u_1(x,t),u_2(x,t)$:
$$u_1(x,t)=U_1(x)b \ \sin(\omega t+\phi)$$

$$u_2(x,t)=U_2(x)b \ \sin(\omega t+\phi)$$
其中:
$$U_1(x)=B_1\sin(\frac{\omega}{a}x)+B_2\cos(\frac{\omega}{a}x)$$

$$U_2(x)=B_3\sin(\frac{\omega}{a}x)+B_4\cos(\frac{\omega}{a}x)$$

(1):
$$U_1(0,t)=0 \implies B_2=0$$
(2):
$$EAU_1^{‘}(s^-)-M{\omega}^2U_1(s^-)=EAU_2^{‘}(s^+)$$

$$\implies
\left[\frac{EA\omega}{a}\cos(\frac{\omega}{a}s)-M{\omega}^2\sin(\frac{\omega}{a}s) \right]B_1-\left[\frac{EA\omega}{a}\sin(\frac{\omega}{a}s)+M{\omega}^2\cos(\frac{\omega}{a}s) \right]B_2-\frac{EA\omega}{a}\cos(\frac{\omega}{a}s)\ B_3+\frac{EA\omega}{a}\sin(\frac{\omega}{a}s)\ B_4=0$$
(3):
$$\frac{EA\omega}{a} \cos(\frac{\omega}{a}L)\ B_3-\frac{EA\omega}{a} \sin(\frac{\omega}{a}L)\ B_4=0$$
(4):
$$\sin(\frac{\omega}{a}s)\ B_1+\cos(\frac{\omega}{a}s)\ B_2-\sin(\frac{\omega}{a}s)\ B_3-\cos(\frac{\omega}{a}s)\ B_4=0$$

将上述四式联立写成矩阵形式:
$$
\begin{pmatrix}
0 & 1 & 0 & 0 \
\left[\frac{EA\omega}{a}\cos(\frac{\omega}{a}s)-M{\omega}^2\sin(\frac{\omega}{a}s) \right] & -\left[\frac{EA\omega}{a}\sin(\frac{\omega}{a}s)+M{\omega}^2\cos(\frac{\omega}{a}s) \right] & -\frac{EA\omega}{a}\cos(\frac{\omega}{a}s) & \frac{EA\omega}{a}\sin(\frac{\omega}{a}s) \
0 & 0 & \frac{EA\omega}{a} \cos(\frac{\omega}{a}L) & -\frac{EA\omega}{a} \sin(\frac{\omega}{a}L) \
\sin(\frac{\omega}{a}s) & \cos(\frac{\omega}{a}s) & -\sin(\frac{\omega}{a}s) & -\cos(\frac{\omega}{a}s) \
\end{pmatrix}
\begin{pmatrix}
B_1 \
B_2 \
B_3 \
B_4 \
\end{pmatrix}
=0
$$

所以要使$B_1,B_2,B_3,B_4 $有不全为0的解，左边矩阵的行列式为0:
$$\implies 1 - \frac{L}{a}\omega \ \tan(\frac{\omega}{a}s) = \tan(\frac{\omega}{a}s) \ \tan(\frac{\omega}{a}L-\frac{\omega}{a}s) $$

取$\frac{L}{a}=10 $,令$\frac{s}{a}=x $:
$$\implies 1 - 10 \omega \ \tan(\omega x) = \tan(\omega x) \ \tan(10 \omega- x \omega) $$

代码:

clc
func_w = @(w,x)(1-10*w.*tan(w.*x)-tan(w.*x).*tan(10.*w-w.*x));
x = linspace(0.01,10,1000);
y = linspace(0,10,1000);
q_1=[];
q_2=[];
q_3=[];
q_4=[];
for i=x
    q = FindZeros(func_w,4,y,i);
    q_1=[q_1 q(1)];
    q_2=[q_2 q(2)];
    q_3=[q_3 q(3)];
    q_4=[q_4 q(4)];
end
figure(1)
plot(x,q_1);
xlabel('\omega_1')
figure(2)
plot(x,q_2);
xlabel('\omega_2')
figure(3)
plot(x,q_3);
xlabel('\omega_3')
figure(4)
plot(x,q_4);
xlabel('\omega_4')