Period of a Pendulum

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Period of a Pendulum

1   Problem

Diagram5Consider 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

Answer:

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}} 

Diagram6From 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

Diagram1Consider 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 .

Answer:

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

Derivation:

Diagram2When 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)} 

Diagram3The 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.