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Vibration refers to the rapid back-and-forth motion of an object around its equilibrium position. This motion can be described in terms of its frequency, amplitude, and direction.
Vibration can occur in a variety of systems, including mechanical systems such as engines and bridges, electronic systems such as speakers and microphones, and even biological systems such as the human body.
Vibration can have both positive and negative effects. For example, it can be used to help mix materials, or to generate electricity in power plants. On the other hand, excessive vibration can cause damage to structures or equipment, or be harmful to human health, such as causing motion sickness or hearing damage.
As a first approximation axial vibration of a space launch vehicle is modelled as a rod of length $L$ and cross-sectional area $A$, with mass per unit length $\rho$ and Young’s modulus $E$. The spatial coordinate $x$ is measured from one end of the rod, and the axial displacement at the location $x$ is $u(x, t)$. During flight a corrective impulsive thrust $I$ is applied at the location $x=x_1$. Just after the impulse (at $t=0$ say) the resulting conditions are $$ u(x, 0)=0, \quad \dot{u}(x, 0)=(I / \rho) \delta\left(x-x_1\right) $$ The rod has free boundary conditions at each end. (a) Assume a solution for the resulting motion of the rod in the form $$ u(x, t)=f(x-c t)+g(x+c t) $$ where $c$ is the compressive wave speed. Write down the initial conditions and boundary conditions that apply to the problem. By focusing on the properties of $f^{\prime}(x)$, rather than $f(x)$, find an expression for $f^{\prime}(x)$ over the range $0 \leq x \leq L$ and show that beyond this range the function is symmetric and has period $2 L$. Hence show that the velocity of the system can be represented as two travelling delta functions which are reflected from the ends of the rod.
We assume a solution for the resulting motion of the rod in the form:
$u(x, t)=f(x-c t)+g(x+c t)$
where $c$ is the compressive wave speed, $f(x-ct)$ is a leftward-travelling wave and $g(x+ct)$ is a rightward-travelling wave.
The initial conditions are:
$\begin{aligned} & u(x, 0)=f(x)+g(x)=0 \ & \dot{u}(x, 0)=c\left(f^{\prime}(x)-g^{\prime}(x)\right)=\frac{I}{\rho} \delta\left(x-x_1\right)\end{aligned}$
The boundary conditions are:
$\begin{aligned} & u(0, t)=f(-c t)+g(c t)=0 \ & u(L, t)=f(L-c t)+g(L+c t)=0\end{aligned}$
By focusing on the properties of $f^{\prime}(x)$, we can write:
$\dot{u}(x, 0)=c\left(f^{\prime}(x)-g^{\prime}(x)\right)=\frac{I}{\rho} \delta\left(x-x_1\right)$
Taking the derivative with respect to $x$ of the above expression and evaluating at $x=x_1$, we obtain:
$\left.f^{\prime}(x)\right|{x=x_1}-\left.g^{\prime}(x)\right|{x=x_1}=\frac{I}{\rho c}$
Solving for $f(x)$, we have:
$f(x)=-g(x)+A \sin \left(\frac{n \pi}{L} x\right)$
where $A$ is a constant determined by the initial conditions and $n$ is an odd integer. Since $f(x)$ is odd and periodic with period $2L$, we have:
$f(x+2 L)=-f(x)$
and
$f^{\prime}(x+L)=-f^{\prime}(x)$
Hence, the velocity of the system can be represented as two travelling delta functions which are reflected from the ends of the rod:
$\dot{u}(x, t)=\frac{I}{2 \rho c}\left[\delta\left(x-x_1-c t\right)+\delta\left(x-x_1+c t\right)\right]+\sum_{n=1}^{\infty} \frac{2 A}{n \pi} \sin \left(\frac{n \pi}{L} x\right) \cos \left(\frac{n \pi c}{L} t\right)$
A cantilever beam of length $L$, uniform cross-section of area $A$ and second moment of area $I$ is made of a material with density $\rho$ and Young’s modulus $E$. (a) The beam is clamped at $x=0$ and free at $x=L$, and undergoes small-amplitude bending vibration with transverse displacement $y(x, t)$. (i) Starting from the governing equation for transverse vibration of a beam, derive an expression for the $n$th mode shape $u_n(x)$ in terms of the wavenumber $k_n$ and the properties of the beam.
The governing equation for the transverse vibration of a beam is given by:
$\frac{\partial^2}{\partial x^2}\left(E I \frac{\partial^2 y}{\partial x^2}\right)+\rho A \frac{\partial^2 y}{\partial t^2}=0$
For the $n$th mode shape, we assume that $y(x,t) = u_n(x)sin(\omega_nt)$, where $\omega_n$ is the $n$th natural frequency. Substituting this into the governing equation and simplifying yields:
$\frac{d^4 u_n}{d x^4}+\frac{\omega_n^2 \rho A}{E I} u_n=0$
This is a fourth-order linear homogeneous differential equation with constant coefficients. The general solution is a linear combination of four functions:
$u_n(x)=A_n \sin \left(k_n x\right)+B_n \cos \left(k_n x\right)+C_n e^{\alpha_n x}+D_n e^{-\alpha_n x}$
where $k_n = \frac{\omega_n}{\sqrt{\frac{EI}{\rho A}}}$ is the wavenumber and $\alpha_n = \frac{\sqrt{\omega_n^2 \rho A / EI}}{2}$ is the decay constant. The constants $A_n$, $B_n$, $C_n$ and $D_n$ can be determined from the boundary conditions.
Find the natural frequencies $\omega_n$ of the beam for the first four modes, using a suitable approximation where needed. Express your answer as factors $\alpha_n$ of the first natural frequency $\omega_1$, i.e. in the form $\omega_n=\alpha_n \omega_1$.
For a cantilever beam, $y(0,t)=0$ and $y'(0,t)=0$, which imply $u_n(0)=0$ and $u_n'(0)=0$. For a free end, $y”(L,t)=0$ and $y”'(L,t)=0$, which imply $u_n”(L)=0$ and $u_n”'(L)=0$.
Using these boundary conditions, we can solve for the natural frequencies and mode shapes. For the first four modes, we have:
Mode 1: $u_1(x)=B_1sin(k_1x)$
At the clamped end, $u_1(0)=0$, which implies $B_1=0$. At the free end, $u_1”(L)=0$, which gives:
$k_1 L=\frac{\pi}{2}$
The first natural frequency is then:
$\omega_1=\frac{\pi}{2 L} \sqrt{\frac{E I}{\rho A}}$
Mode 2: $u_2(x)=C_2e^{\alpha_2 x}+D_2e^{-\alpha_2 x}$
At the clamped end, $u_2(0)=0$, which implies $C_2+D_2=0$. At the free end, $u_2”(L)=0$, which gives:
$\alpha_2^4 L^4-\alpha_2^2 L^2+k_2^2=0$
This equation cannot be solved analytically, but we can use a suitable approximation where $\alpha_2 L \ll 1$. In this case, the above equation reduces to:
$k_2 L=\frac{\pi}{L}$
The second natural frequency is then:
$\omega_2=\frac{\pi}{L} \sqrt{\frac{2 E I}{\rho A}}$
Mode 3: $u_3(x)=B_3cos(k_3x)$
At the clamped end, $u_3
