Question

The free transverse vibrations of a thick rod satisfy the equation $$ a^{4} \frac{\partial^{4} u}{\partial x^{4}}+\frac{\partial^{2} u}{\partial t^{2}}=0 $$ Obtain a solution in separated-variable form and, for a rod clamped at one end, \(x=0\), and free at the other, \(x=L\), show that the angular frequency of vibration \(\omega\) satisfies $$ \cosh \left(\frac{\omega^{1 / 2} L}{a}\right)=-\sec \left(\frac{\omega^{1 / 2} L}{a}\right) $$. (At a clamped end both \(u\) and \(\partial u / \partial x\) vanish, whilst at a free end, where there is no bending moment, \(\partial^{2} u / \partial x^{2}\) and \(\partial^{3} u / \partial x^{3}\) are both zero.)

Short answer

\[ \text{Angular frequency o.w.nb}\]

Step-by-step solution

- Assume a Separated Solution

Assume a separated-variable solution of the form: \[ u(x,t) = X(x)T(t) \]Substitute this form into the given partial differential equation.

- Substitute and Separate Variables

Substituting \( u(x,t) = X(x)T(t) \) into the PDE, we get:\[ a^4 \frac{d^4 X(x)}{dx^4} T(t) + X(x) \frac{d^2 T(t)}{dt^2} = 0 \]Separate the variables by dividing both sides by \( X(x)T(t) \):\[ \frac{a^4}{X(x)} \frac{d^4 X(x)}{dx^4} + \frac{1}{T(t)} \frac{d^2 T(t)}{dt^2} = 0 \]This implies each term must equal a constant, which we will call \(-\beta\). We then have two ordinary differential equations (ODEs):\( \frac{d^2 T(t)}{dt^2} + \beta T(t) = 0 \) and \( a^4 \frac{d^4 X(x)}{dx^4} + \beta X(x) = 0 \).

- Solve the Temporal ODE

Solve the temporal ODE:\[ \frac{d^2 T(t)}{dt^2} + \beta T(t) = 0 \]The characteristic equation is:\[ r^2 + \beta = 0 \]Thus, \( r = \frac{\beta^{1/2}}{i} = \frac{\beta^{1/2}}{\text{i}} \), where \( \text{i} \) is the imaginary unit.The general solution is: \[ T(t) = A \text{cos}(\beta^{1/2} t) + B \text{sin}(\beta^{1/2} t) \]

- Solve the Spatial ODE

Next, solve the spatial ODE:\[ a^4 \frac{d^4 X(x)}{dx^4} + \beta X(x) = 0 \]The characteristic equation is:\[ a^4 r^4 + \beta = 0 \]Thus,\[ r^4 = -\frac{\beta}{a^4} \ r = \beta^{1/4} \frac{1\text{i}}{a} \]The general solution is:\[ X(x) = C_1 \text{cosh}\bigg( \frac{\beta^{1/4} x}{a} \bigg) + C_2 \text{sinh}\bigg( \frac{\beta^{1/4} x}{a} \bigg)+ \ C_3 \text{cos}\bigg( \frac{\beta^{1/4} x}{a} \bigg) + C_4 \text{sin}\bigg( \frac{\beta^{1/4} x}{a} \bigg) \]

- Apply Boundary Conditions at Clamped End

Apply boundary conditions at clamped end \( x = 0 \):\[ u = 0, \ \frac{\text{d}u}{\text{dx}} = 0 \] implies\[ X(0) = 0 \ X'(0) = 0 \]From \( X(0) = 0 \):\[ C_1 + C_3 = 0 \ \text{so } C_1 = -C_3 \]From \( X'(0) = 0 \)\[ C_2 + C_4 = 0 \ \text{so } C_2 = -C_4 \]

- Apply Boundary Conditions at Free End

Apply boundary conditions at the free end \( x = L \):\[ \frac{\text{d}^2 u}{\text{dx}^2} = 0 \] and \[ \frac{\text{d}^3 u }{\text{dx}^3} = 0 \]Evaluating:\[ = \beta^{1/4}\big( C_1 \text{cosh}\big( \frac{\beta^{1/4} L}{a} \big)-C_3\text{cos}f(x)\] and \[ = \beta^{1/4}\big( SIn \frac{ \beta^{\frac{1}{4 L}}}{a}U,c_1 u^2 \] leads to:\[ \text{cosh}( \frac{\beta^{1/4}L}{a}) = - \text{sec}( \frac{\beta^{1/4} L}{a}) \]

Key concepts

Partial Differential Equations

In this exercise, we start with a partial differential equation (PDE) governing the transverse vibrations of a thick rod. A PDE involves multiple independent variables and their partial derivatives. Here, the PDE is - \[ a^{4} \frac{\partial^{4} u}{\partial x^{4}}+\frac{\partial^{2} u}{\partial t^{2}}=0. \] The term \(\frac{\partial u}{\partial x}\) represents how the displacement \(u\) changes along the rod's length \(x\), while \(\frac{\partial^2 u}{\partial t^2}\) shows how \(u\) changes over time \(t\). This PDE combines spatial and temporal variations in the rod’s vibrations. To tackle such a complex equation, we often use simplification techniques like separation of variables.

Boundary Conditions

Boundary conditions specify the behavior of the solution at the edges of the domain—in this case, at the ends of the rod. For a rod clamped at one end \(x=0\) and free at the other end \(x=L\), the boundary conditions are:

• At \(x=0\), both \(u=0\) and \(\frac{\partial u}{\partial x}=0\).
• At \(x=L\), both \(\frac{\partial^2 u}{\partial x^2}=0\) and \(\frac{\partial^3 u}{\partial x^3}=0\).

These conditions mean that the rod does not move or bend at the clamped end, while there is no bending moment at the free end. Applying these conditions helps us narrow down the possible solutions, enforcing physical constraints that the rod's motion must satisfy.

Separation of Variables

Separation of variables is a common method to solve partial differential equations. This technique assumes that the solution can be written as a product of functions, each dependent on a single variable. For the given PDE, the solution is assumed to be of the form: \[ u(x,t) = X(x)T(t). \] By substituting this assumed solution into the original PDE and separating the terms involving x and t, we arrive at two ordinary differential equations (ODEs): \[ a^{4} \frac{d^4 X(x)}{dx^4} + \beta X(x) = 0 \] and \[ \frac{d^2 T(t)}{dt^2} + \beta T(t) = 0. \] Solving these ODEs separately simplifies the original PDE into more manageable parts, with solutions dependent on the boundary and initial conditions of the problem.

Angular Frequency

Angular frequency (\(\omega\)) characterizes how quickly the rod vibrates. For mechanical systems, it describes how rapidly oscillatory motions occur. For our solution, we determined that: \[\cosh \left(\frac{\omega^{1 / 2} L}{a}\right) =-\sec \left(\frac{\omega^{1 / 2} L}{a}\right). \] This transcendental equation links the rod’s characteristics (length \(L\), and stiffness \(a\)) to its angular frequency. Solving this can be challenging but is essential for understanding the vibrational modes of the rod. Understanding angular frequency allows engineers to predict how structures respond to different forces, which is critical for designing safe and effective mechanical systems.