harmonic-oscillatoroscillators
I have been looking for the derivation/place where I can quote the formula $T=\frac{2\pi}{r^2}\sqrt{\frac{2IL}{\eta\pi}}$ from as I remember seeing in class but can't seem to find it online to quote/derive it for my self.
Clarification:
For
$$T=\frac{2\pi}{r^2}\sqrt{\frac{2IL}{\eta\pi}}$$
$I$ is the moment of inertia, $L$ is suspension length, $\eta$ is the rigidity modulus of the material and $T$ is the time period.
Any help would be greatly appreciated.
You can find in Sakurai this expression for the propagator $$K({\bf{x''}},t;{\bf{x'}},t) = \sum\limits_{a'} {\left\langle {x''\left| {a'} \right\rangle } \right.} \left\langle {a'\left| {x'} \right\rangle } \right.\exp \left[ {\frac{{ - i{E_{a'}}(t - {t_0})}}{\hbar }} \right]$$
Taking into account the QHO wavefunctions
$${a_n}(x) = \frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4}} \cdot {e^{ - \frac{{m\omega {x^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x} \right),\qquad n = 0,1,2, \ldots . $$
and
$$\left\langle {x''\left| {a'} \right\rangle } \right. = \frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4}} \cdot {e^{ - \frac{{m\omega x'{'^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right), $$
We simply can derive the expression for $K({\bf{x''}},t;{\bf{x'}},t)$
$$\begin{array}{l}K({\bf{x''}},t;{\bf{x'}},t) = \sum\limits_{a'} {\frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {{\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)}^{1/4}} \cdot {e^{ - \frac{{m\omega x'{'^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)} \times \\ \times \frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4}} \cdot {e^{ - \frac{{m\omega x{'^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right)\exp \left[ {\frac{{ - i{E_{a'}}(t - {t_0})}}{\hbar }} \right] = \\\frac{1}{{{2^n}{\mkern 1mu} n!}} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\sum\limits_{n = 0} {{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\exp \left[ { - i\omega (n + 1/2)(t - {t_0})} \right]} = \\ = \frac{1}{{{2^n}{\mkern 1mu} n!}} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\sum\limits_{n = 0} {{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\exp \left[ { - i\omega (n + 1/2)(t - {t_0})} \right]} = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}{e^{ - \frac{{i\omega }}{2}(t - {t_0})}}{\sum\limits_{n = 0} {\frac{1}{{{2^n}{\mkern 1mu} n!}}{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\left( {{e^{ - i\omega (t - {t_0})}}} \right)} ^n} = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}{e^{ - \frac{{i\omega }}{2}(t - {t_0})}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{\hbar }}}{\sum\limits_{n = 0} {\frac{1}{{{2^n}{\mkern 1mu} n!}}{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\left( {{e^{ - i\omega (t - {t_0})}}} \right)} ^n} = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}{e^{ - \frac{{i\omega }}{2}(t - {t_0})}}\left[ {\frac{1}{{\sqrt {1 - {e^{ - 2i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{x{'^2} + x'{'^2} - 2x''x'{e^{ - i\omega (t - {t_0})}}}}{{1 - {e^{ - 2i\omega (t - {t_0})}}}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{\sqrt {1 - {e^{ - 2i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{{e^{i\omega (t - {t_0})}} - {e^{ - i\omega (t - {t_0})}}}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{{e^{\frac{{i\omega }}{2}(t - {t_0})}}\sqrt {1 - {e^{ - 2i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{\sqrt {{e^{i\omega (t - {t_0})}} - {e^{ - i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{m\omega }}{{2\hbar }}\left[ {\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{i\sin (\omega (t - {t_0}))}} - (x{'^2} + x'{'^2})} \right]} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{im\omega }}{{2\hbar }}\frac{{(x{'^2} + x'{'^2})\cos (\omega (t - {t_0})) - 2x''x'}}{{\sin (\omega (t - {t_0}))}}} \right)\\\end{array} $$
The last moment you can see the Hermitian polinomial functions is when we apply your formula
This bothered me as an undergraduate too, and it wasn't until much later that I found out the rigorous mathematical way of understanding perturbation theory. The basic answer is that "we are restricting our attention to perturbations with this property." For the details, my answer will be loosely cribbed from Wald's General Relativity, which gives a better mathematical description of what we're really doing here than most classical mechanics texts do.
Let's suppose we want to solve a differential equation $\mathcal{E}[q_i(t)] = 0$, where $\mathcal{E}$ stands for some non-linear operator on the functions $q_i(t)$. Let us assume that there exists a family of exact solutions $q_i(t; \lambda)$ to the equations of motion, parameterized by a parameter $\lambda$, with the following properties:
In some sense, $\lambda$ measures the "size" of the perturbation away from the background solution. In particular, since all of our solutions are exact, we can say that $$ \left.\frac{d}{d\lambda} \mathcal{E}[q_i(t; \lambda)] \right|_{\lambda = 0} = 0, $$ and it is not too hard to see that this equation will be a linear equation in the functions $$ \gamma_{i}(t) \equiv \left. \frac{d q_i(t;\lambda)}{d\lambda}\right|_{\lambda = 0}. $$ For sufficiently small $\lambda$, the quantity $q_{0i}(t) + \lambda \gamma_i(t)$ will be a good approximation to $q_i(t;\lambda)$, allowing us to study solutions that are "close" to our background solution. The $\eta_i(t)$ used by Goldstein would be equal to $\lambda \gamma_i(t)$ in this language.
Once you have put all of this into place, then it is fairly straightforward to show that $\dot{\eta}_i(t)$ must go to zero as $\lambda \to 0$, since $$ \frac{d \eta_i}{dt} = \frac{d}{dt} \left[ \lambda \left. \frac{d q_i(t, \lambda)}{d \lambda} \right|_{\lambda = 0}\right] = \lambda \left.\frac{d^2 q_i(t,\lambda)}{dt d\lambda} \right|_{\lambda = 0} $$ and the smoothness assumption on the family $q_i(t;\lambda)$ ensures that this second derivative exists.
To see why the smoothness assumption is what saves us here from the pathology you're considering, consider the one-parameter family of functions $$ f(t; \lambda) = \lambda \sin (t/\lambda). $$ It is certainly the case that as $\lambda \to 0$, we have $\eta \to 0$ but $\dot{\eta} \not\to 0$. But this family of functions is not smooth in $\lambda$, since $$ \frac{df}{d\lambda} = \sin(t/\lambda) - \frac{t}{\lambda}\cos(t/\lambda) $$ and this is not well-defined at $\lambda = 0$. The assumption of a "smooth family of solutions" means that we have eliminated such pathology in the first place, though, and so we don't have to worry about such cases when we're trying to linearize our equations.
Best Answer
For relatively small (see Note) angular twists of the wire, the torque $\tau$ which restores the wire to its untwisted position is proportional to the angle $\theta$ through which the end of the wire is rotated : $$\tau=-\kappa \theta$$
$\kappa$ is called the torsion constant of the wire. The minus sign indicates that the direction of the torque is opposite to the direction in which angle $\theta$ is increasing.
The torque causes rotational acceleration $\ddot\theta$ of mass at the end of the wire. Newton's 2nd Law for this acceleration is $$\tau=I\ddot\theta$$ where $I$ is the moment of inertia of the mass. This is the rotational equivalent of $F=ma$.
Combining the above two equations, the equation of motion for the torsional pendulum is $$\ddot\theta+\frac{\kappa}{I}\theta=0$$ This has the form $\ddot x+\omega^2 x=0$ which describes Simple Harmonic Motion. Here $\omega=2\pi f$ is the angular frequency of the periodic motion (radians per second) and $f$ is frequency (cycles per second; one cycle is $2\pi$ radians). Period and frequency are related by $f=\frac{1}{T}$ and $\omega=\sqrt{\frac{\kappa}{I}}$ therefore $$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{I}{\kappa}}$$
This can be compared with the period of a mass on a spring : $$T=2\pi\sqrt{\frac{m}{k}}$$ The moment of inertia $I$ is equivalent to mass $m$ and the torsion constant $\kappa$ is equivalent to the force constant $k$ or stiffness of the spring.
The final difficult step is to relate torsion constant $\kappa$ to the dimensions of the wire and its shear modulus $\eta$. This calculation is done in Deriving the Shear Modulus S From the Torsion Constant κ. The result is $$\kappa=\frac{\eta \pi r^4}{2L}$$ in which $r$ is the radius of the wire and $L$ its length. Substituting into the equation for the period we get $$T=2\pi\sqrt{\frac{2LI}{\eta \pi r^4}}$$ which is the formula you were given.
Note
The angle $\theta$ through which the end of the wire is rotated is not the same as the angle of twist $\psi$ along the length of the wire. Whereas $\theta$ is measured by the rotation of a radius in the base of the wire, $\psi$ is measured by the twist of a line parallel to the axis. The two are related by $$L\psi=r\theta$$
To ensure that the elastic limit of the wire is not exceeded $\psi$ should be small, typically no more than $10^{\circ}$. However since $L\gg r$ the angle $\theta$ can be quite large. The base can be rotated a whole circle without exceeding the elastic limit.