# Why the time period cannot depend upon the angular velocity and angular acceleration of the pendulum?

To derive the time period $T$ of a pendulum using dimensional analysis it is assumed that it depends upon the mass $m$ of the bob, the length of the string $\ell$, the acceleration due to gravity $g$ and the amplitude $\theta_0$. Then, using dimensional analysis, we find $$T=f(\theta_0)\sqrt{\frac{\ell}{g}}$$ where $f(\theta_0)$ is some unknown dimensionless function the dimensionless parameter $\theta_0$.

1. How can we a priori rule out that $T$ will not depend on the instantaneous values of angular position $\theta(t)$, angular velocity $\dot{\theta}(t)$ and angular acceleration $\ddot{\theta}(t)$? What forbids $T$ to depend upon both $\dot{\theta}(t)$ and $\ddot{\theta}(t)$ in such a way that the time dependences cancel out?

1. If $T$ depends only on the constant values, why not initial velocity $\dot{\theta}(t_0)$ and $\ddot{\theta}(t_0)$?

I'm trying to argue physically without solving the equation of motion.

• $\theta_0$ is an amplitude that does not (in the undamped case) depend on time while, of course, the angular position $\theta(t)$ does depend on time as does $\omega(t) = \dot\theta(t)$ and $\alpha(t) = \ddot\theta(t)$. Why then did you (1) not ask about dependence on $\omega_0$ and $\alpha_0$ in accord with $\theta_0$ or (2) not ask about $\theta(t)$ in accord with $\omega(t)$ and $\alpha(t)$? – Alfred Centauri Oct 19 '17 at 1:56
• @AlfredCentauri Agreed. I've modifed the question. – mithusengupta123 Oct 19 '17 at 3:07

## 3 Answers

I think you wanted to consider a dependence on the amplitude $\dot\theta_0$ of the angular speed and on the amplitude $\ddot\theta_0$ of the angular acceleration, since you can only consider constants as pointed out in @nasu's answer. So let's do that:

$$T = m^\alpha l^\beta g^\gamma \dot\theta_0^\epsilon \ddot\theta_0^\eta.$$

So plugging the dimentions,

$$[T] \propto [M]^\alpha[L]^\beta[LT^{-2}]^\gamma[T^{-1}]^\epsilon[T^{-2}]^\eta,$$

which implies that

\begin{align} \alpha&=0,\\ \beta+\gamma&=0,\\ -2\gamma-\epsilon-2\eta&=1. \end{align}

As you see, you don't have enough equations to nail down all the exponents. The best conclusion you can reach is that

$$T\propto \left(\frac{l}{g}\right)^{\frac{1}{2}+\eta+\frac{1}{2}\epsilon}\dot\theta_0^\epsilon \ddot\theta_0^\eta.$$

You reach the limits of dimensional analysis here.

Well, the EOM from Newton's 2nd law is a 2nd order ODE, i.e. it has 2 integration constants, and we have already specified 2 initial conditions: $$\theta(t_0)~=~\theta_0\qquad\text{and}\qquad\dot{\theta}(t_0)~=~0,$$ cf. the definition of the amplitude $\theta_0$. (It cannot depend on the initial time $t_0$ because of time translation symmetry.)

• What if instead of starting the particle from one of the extreme positions, we set the initial time $t=t_0$ when the particle passes through the equilibrium position $\theta=0$? At that time $\dot{\theta}(t_0)\neq 0$. @Qmechanic – mithusengupta123 Oct 19 '17 at 3:42

Dimensional analysis seeks to understand relations between constants in your problem: these are $m$, $g$, $\ell$ and the initial position $\theta_0$. The basic idea is that this constant captures something “natural” about the system, independent of specific details of the problem. Since $\dot\theta$ or $\ddot{\theta}$ are not constants, they are not considered. Thus, the Planck units, for instance, are constants obtained in term of various fundamental constants of Nature.

One way to understand this is to rephrase your equation as the search for a dimensionless constant quantity $\Pi$ of the form: $$\Pi= \frac{T}{f(\theta_0)} \ell^{a}g^b m^d = \frac{T}{f(\theta_0)} \sqrt{\frac{g}{\ell}}\, . \tag{1}$$ with $f(\theta_0)$ dimensionless. Extending (1) to functions of $\dot\theta$ or $\ddot{\theta}$ would imply that $\Pi$ is not a constant, contrary to the basic hypothesis.

Dimensional analysis says nothing about the function $f(\theta_0)$, which must be determined from experiment or from some more detailed theory. It does tell us that, if you start your pendulum always as $\theta_0$ and increase $\ell$ by $4$ the period $T$ will increase by $2$, which is true irrespective of any other details.

• "Dimensional analysis seeks to understand relations between constants in your problem" Is this statement always true? Dimensional analysis can be used to guess Ohm's law $V=IR$ (where $V$ and $I$ are not constants) or Joule heating law $Q\propto I^2Rt$ or Stokes' law of viscosity $F\propto \eta rv$ and many more. – mithusengupta123 Oct 19 '17 at 3:04
• In $T\sim \sqrt{\ell/g}$, $\ell$ does not vary in time and is considered constant, i.e. for some given $\ell$ what is the relation between this $\ell$ the period $T$ and other constants in the problem. In the example of your ideal resistor, the resistance is considered constant, and you are seeking a functional relation between some constant current, voltage and resistance. – ZeroTheHero Oct 19 '17 at 3:47