Develop the theory of the normal-mode torsional vibrations of a rod of length \(l\) having free ends. Show how to express an arbitrary vibration as a sum of normal-mode vibrations. Show how to evaluate the normal-mode amplitudes from an initial configuration \(\phi(x)\) and an initial velocity \(\partial \phi(x) / \partial t\).

We begin by deriving the equation governing the torsional vibrations of a rod with free ends. For a rod of length \(l\) subject to torsional vibrations, we have the equation of motion: \(\frac{\partial^2 \phi(x,t)}{\partial t^2} = c^2 \frac{\partial^2 \phi(x,t)}{\partial x^2}\) where \(\phi(x,t)\) is the angular displacement of the rod at position \(x\) and time \(t\) and \(c\) is the speed of torsional waves in the material. This is a partial differential equation that can be solved by separation of variables.

Next, we need to solve the equation of motion for the normal modes. To do this, we introduce a separation of variables ansatz: \(\phi(x,t) = X(x) T(t)\) Substituting this ansatz into the governing equation of motion yields: \(X(x) \frac{d^2 T(t)}{dt^2} = c^2 T(t) \frac{d^2 X(x)}{dx^2}\) Dividing both sides by \(c^2 XT\), we get: \(\frac{1}{c^2 T(t)} \frac{d^2 T(t)}{dt^2} = \frac{1}{X(x)} \frac{d^2 X(x)}{dx^2}\) Since the left side depends only on time and the right side depends only on position, they must be equal to a constant. We denote this constant as \(-\omega^2\): \(\frac{1}{c^2 T(t)} \frac{d^2 T(t)}{dt^2} = \frac{1}{X(x)} \frac{d^2 X(x)}{dx^2} = -\omega^2\) Solving these two ordinary differential equations with the conditions \(X(0)=X(l)=0\), we find the normal modes: \(X_n(x) = \sin(\frac{n \pi x}{l})\) \(T_n(t) = \cos(\omega_n t)\) where the index \(n\) enumerates the modes and the frequencies \(\omega_n\) are given by: \(\omega_n = c \frac{n \pi}{l}\)

We can now express an arbitrary torsional vibration as a sum of normal-mode vibrations: \(\phi(x,t) = \sum_{n=1}^\infty A_n \sin(\frac{n \pi x}{l}) \cos(\omega_n t)\) where the \(A_n\)s are the normal-mode amplitudes.

Finally, we need to evaluate the normal-mode amplitudes given the initial configuration \(\phi(x)\) and initial velocity \(\frac{\partial \phi(x)}{\partial t}\). At \(t=0\), we have: \(\phi(x) = \sum_{n=1}^\infty A_n \sin(\frac{n \pi x}{l})\) and \(\frac{\partial \phi(x)}{\partial t} = \sum_{n=1}^\infty A_n \omega_n \sin(\frac{n \pi x}{l})\) To evaluate the normal-mode amplitudes, we can use the orthogonality of the sine functions and integrate both sides over the interval \([0, l]\): \(A_n = \frac{2}{l} \int_0^l \phi(x) \sin(\frac{n \pi x}{l}) dx\) In conclusion, this is how we develop the theory of normal-mode torsional vibrations of a rod of length \(l\) having free ends. We show how to express an arbitrary vibration as a sum of normal-mode vibrations and how to evaluate the normal-mode amplitudes from an initial configuration \(\phi(x)\) and an initial velocity \(\frac{\partial \phi(x)}{\partial t}\).