Consider free torsional vibration of a uniform circular elastic rod of length \(L\), torsional stiffness \(G J\) and mass per unit length \(m\), that is free at its left end and fixed at its right end. (a) Establish the modal boundary conditions for the structure. (b) Derive the frequency equation for the rod. (c) Determine the first three natural frequencies and modal functions. Plot the first three modes.

Since the rod is fixed at the right end, the angular displacement at this point should be zero, i.e., \(\theta(L) = 0\). Meanwhile, the left end is free, the torque at the left end should be zero, which means the derivative of angular displacement with respect to the position should be zero at the left end: \(\frac{d\theta(0)}{dx} = 0\). These boundary conditions are essential for solving the frequency equation in the next step.

To derive the frequency equation for the rod, we need to consider the torsional vibration equation: \(\frac{d^2\theta(x)}{dx^2} + \frac{m}{G J}\omega^2\theta(x) = 0\). Here, \(\omega\) is the natural frequency of torsional vibration. We can rewrite this equation and insert the boundary conditions from Step 1: \(\frac{d^2\theta}{dx^2} + k^2 \theta = 0,\;\;\; k^2 = \frac{m\omega^2}{G J}\), with \(\theta(0) = 0\) and \(\frac{d\theta(L)}{dx} = 0\). Now we have an ordinary differential equation (ODE) to solve and find the required frequency equation. The general solution to the ODE above can be written as: \(\theta(x) = A\sin(kx) + B\cos(kx)\). Now, we will apply the boundary conditions: 1. For a free end: \(\theta(0) = 0\), then \(B = 0\). 2. For a fixed end: \(\frac{d\theta(L)}{dx} = 0\), we have \(A k\cos(kL) = 0\). Therefore, the second boundary condition implies that \(kL\) must be equal to odd integer multiples of \(\frac{\pi}{2}\): \(kL = (2n + 1) \frac{\pi}{2}\), with \(n = 0, 1, 2, ...\). Finally, we obtain the frequency equation for the torsional vibration of a rod: \(\omega_n = \frac{(2n+1)\pi}{2L}\sqrt{\frac{G J}{m}}\), with \(n = 0, 1, 2, ...\).

Now that we have the frequency equation, we can easily calculate the first three natural frequencies by substituting \(n = 0, 1, 2\): 1. For \(n = 0\): \(\omega_1 = \frac{\pi}{2L}\sqrt{\frac{G J}{m}}\). 2. For \(n = 1\): \(\omega_2 = \frac{3\pi}{2L}\sqrt{\frac{G J}{m}}\). 3. For \(n = 2\): \(\omega_3 = \frac{5\pi}{2L}\sqrt{\frac{G J}{m}}\). The corresponding modal functions for each frequency are: 1. For \(\omega_1\): \(\theta_1(x) = A_1\sin(\frac{\pi x}{2L})\). 2. For \(\omega_2\): \(\theta_2(x) = A_2\sin(\frac{3\pi x}{2L})\). 3. For \(\omega_3\): \(\theta_3(x) = A_3\sin(\frac{5\pi x}{2L})\). The plot of the first three modes can be shown by illustrating \(\theta_1(x)\), \(\theta_2(x)\), and \(\theta_3(x)\) for several periods. The plots display the torsional displacement versus the rod's position for different natural frequencies. Note that the amplitudes \(A_1, A_2, A_3\) solely affect the amplitudes of the modes and can be chosen arbitrarily for plotting purposes.