# Period of a Pendulum

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Period of a Pendulum

1   Problem

Consider a block of mass $m$ released at the top of a frictionless quarter-circle incline of radius $R$. How much time $t$ does it take the block to travel to the bottom? Assume the dimensions of the box are much smaller than $R$ and that the block stays in contact with the surface of the incline at all times. The acceleration due to gravity is given by $g$.

You may make use of the integral

$\alpha \equiv \int_0^{\pi/2} \frac{d\theta}{\sqrt{\cos\theta}} \approx 2.62206$

and use $\alpha$ in your solution.

2   Solution

$t = \alpha\sqrt{\frac{R}{2g}}$

Derivation:

Look at the general case described below. Here we will use:

$f(x) = R - \sqrt{R^2-x^2}$

$f'(x) = \frac{x}{\sqrt{R^2-x^2}}$

$f^{-1}(x) = \sqrt{2Rx - x^2}$

$h = R$

$t = \int_0^{f^{-1}(h)} \sqrt{\frac{1+f'^2(x)}{2g(h-f(x))}}dx$

$f^{-1}(R) = R$

$1 + f'^2(x) = \frac{R^2}{R^2-x^2}$

$2g(h-f(x)) = 2g\sqrt{R^2-x^2}$

$\frac{1 + f'^2(x)}{2g(h-f(x))} = \frac{R^2}{2g(R^2-x^2)^{3/2}}$

$t = \int_0^R \sqrt{\frac{R^2}{2g(R^2-x^2)^{3/2}}}dx$

$t = \frac{R}{\sqrt{2g}} \int_0^R \frac{dx}{(R^2-x^2)^{3/4}}$

From the diagram,

$\sin\theta = \frac{x}{R}$

$dx = R\cos\theta d\theta$

$R^2-x^2 = R^2\cos^2\theta$

$x = 0, \theta = 0$

$x = R, \theta = \pi/2$

$t = \frac{R}{\sqrt{2g}} \int_0^{\pi/2} \frac{R\cos\theta d\theta}{(R\cos\theta)^{3/2}}$

$t = \frac{R}{\sqrt{2g}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{R\cos\theta}}$

$t = \sqrt{\frac{R}{2g}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{\cos\theta}}$

3   Pendulums

The motion of a block down a circular incline is identical to that of a simple pendulum. In this case, the period of a simple pendulum of varying amplitude can be derived by changing the height $h$ from which the block is released.

$f^{-1}(h) = \sqrt{2Rh - h^2}$

$2g(h-f(x)) = 2g(h - R + \sqrt{R^2-x^2})$

$\frac{1 + f'^2(x)}{2g(h-f(x))} = \frac{R^2}{2g(R^2-x^2)(h-R+\sqrt{R^2-x^2})}$

$t = \int_0^{\sqrt{2Rh - h^2}} \sqrt{\frac{R^2}{2g(R^2-x^2)(h-R+\sqrt{R^2-x^2})}}dx$

$t = \frac{R}{\sqrt{2g}} \int_0^{\sqrt{2Rh - h^2}} \frac{dx}{\sqrt{(R^2-x^2)(h-R+\sqrt{R^2-x^2})}}$

Putting $h$ in terms of the amplitude $\theta$, it can be seen that $R - h = R\cos\theta$. Therefore, $h = R(1-\cos\theta)$.

$\sqrt{2Rh - h^2} = R\sin\theta$

$t = \frac{R}{\sqrt{2g}} \int_0^{R\sin\theta} \frac{dx}{\sqrt{(R^2-x^2)(\sqrt{R^2-x^2}-R\cos\theta)}}$

Period of a Pendulum of Amplitude $\theta_o$

$T = \sqrt{\frac{8L}{g}} \int_0^{\theta_o} \frac{d\theta}{\sqrt{(\cos\theta-\cos\theta_o)}}$

Derivation:

$T = 4t$

$T = \frac{4R}{\sqrt{2g}} \int_0^{R\sin\theta_o} \frac{dx}{\sqrt{(R^2-x^2)(\sqrt{R^2-x^2}-R\cos\theta_o)}}$

Using the previous diagram,

$x = 0, \theta = 0$

$x = R\sin\theta_o, \theta = \theta_o$

$T = \frac{4R}{\sqrt{2g}} \int_0^{\theta_o} \frac{R\cos\theta d\theta}{\sqrt{(R^2\cos^2\theta)(R\cos\theta-R\cos\theta_o)}}$

$R=L$

$T = \sqrt{\frac{8L}{g}} \int_0^{\theta_o} \frac{d\theta}{\sqrt{(\cos\theta-\cos\theta_o)}}$

4   General Case

Consider a block of mass $m$ released at a height $h$ on a frictionless, curved incline described by the function $y = f(x)$. How much time $t$ does it take the block to travel to the bottom? Assume the dimensions of the box are much smaller than $h$ and that the block stays in contact with the surface of the incline at all times. The acceleration due to gravity is given by $g$.

$t = \int_0^{f^{-1}(h)} \sqrt{\frac{1+f'^2(x)}{2g(h-f(x))}}dx$

Derivation:

When the block is located at a height $y$, its velocity can be derived from conservation of energy.

$mg(h-y) = \frac{1}{2}mv^2$

$v = \sqrt{2g(h-y)}$

The time $dt$ it takes the block to travel down the incline is related to the distance $dl$ given by:

$dl = \frac{v_o+v_f}{2}dt$

From the diagram:

$dl = \sqrt{(dx)^2+(dy)^2}$

From above, $v_o$ is given by $\sqrt{2g(h-y)}$. $v_f$ is given by $\sqrt{2g(h-y+dy)}$. Ignoring first order differentials, $v_o = v_f = \sqrt{2g(h-y)}$.

$\sqrt{(dx)^2+(dy)^2} = \sqrt{2g(h-y)}dt$

$dy = f'(x) dx$

$\sqrt{1 + f'^2(x)}dx = \sqrt{2g(h-f(x))}dt$

$\sqrt{\frac{1+f'^2(x)}{2g(h-f(x))}}dx = dt$

$\int_0^{f^{-1}(h)} \sqrt{\frac{1+f'^2(x)}{2g(h-f(x))}}dx = \int_0^t dt'$

$t = \int_0^{f^{-1}(h)} \sqrt{\frac{1+f'^2(x)}{2g(h-f(x))}}dx$

Restrictions on $f(x)$:

We will require $f''(x) \geq 0$ in our interval so that the block’s velocity vector never points above the surface and hence the block will stay on the incline.

Simplifications:

The solution

$t = \int_0^{f^{-1}(h)} \sqrt{\frac{1+f'^2(x)}{2g(h-f(x))}}dx$

gives the exact expression for the time taken, however no elementary antiderivative exists for most functions. We will make a practical approximation that will make it easy to evaulate the integral. If $f'(x) \gg 1$, we can rewrite $t$ as:

$t \approx \int_0^{f^{-1}(h)} \sqrt{\frac{f'^2(x)}{2g(h-f(x))}}dx$

$t \approx \frac{1}{\sqrt{2g}}\int_0^{f^{-1}(h)} \frac{f'(x) dx}{\sqrt{(h-f(x))}}$

We can make the u-substitution $u = h - f(x)$ and $du = - f'(x) dx$.

$-\int \frac{du}{\sqrt{u}} = -2\sqrt{u}+C$

$\int \frac{f'(x) dx}{\sqrt{(h-f(x))}} = -2\sqrt{h-f(x)} + C$

$\int_0^{f^{-1}(h)} \frac{f'(x) dx}{\sqrt{(h-f(x))}} = -2\sqrt{h-f(f^{-1}(h))} + 2\sqrt{h-f(0)}$

$t \approx \frac{2\sqrt{h-f(0)}}{\sqrt{2g}}$

If we have $f(0) = 0$, then

$t \approx \sqrt{\frac{2h}{g}} = t_o$

which is the time it takes for the block to fall a height $h$ when released from rest. We can see that $t_o$ is sufficient in most cases.