# How is the time period of a pendulum indepedent of the amplitude provided?The time period of the pendulum remains the same if released from a greater amplitude or a shorter amplitude. How can this...

How is the time period of a pendulum indepedent of the amplitude provided?

The time period of the pendulum remains the same if released from a greater amplitude or a shorter amplitude. How can this be proved?

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### 2 Answers

You can test this experimentally yourself. Set up a pendulum of known length and vary the angle from vertical in small increments. Measure the angle and determine the length of the period - the time it takes to make one swing back and forth. Then graph the period vs. the angle. The period presumably depends on the angle so put period on y-axis and angle on x-axis. Draw a line through your points. If period is independent of angle, you should have a straight, horizontal line on your graph.

In general, period is independent only at small angles.

It is not true. The period of a pendulum is always dependent on the amplitude, but is approximately true for small amplitudes where sin(theta) is almost theta.

In these equations theta must be given in radians, and for this to work theta must be much less than 1 radian.

The force on the pendulum is -mg sin(theta)

s = len*theta

ds/dt = len * d(theta)/dt

d^2(s)/dt^2 = len * d^2(theta)/dt^2 so

F = ma = m * len * d^2(theta)/dt^2 and our differential equation is

m (len * d^2(theta)/dt^2) = -mg sin(theta) solving we get

d^2(theta)/dt^2 + g/l sin(theta) = 0

This equation cannot be solved in elementary functions, but if we use sin(theta) is approximately theta

the differential equation

d^2(theta)/dt^2 + g/l (theta) = 0

gives a solution

theta(t) = theta(0) cos(sqrt(g/l) t)

The period of such an equation is 2pi/(sqrt(g/l)) or

T = 2pi sqrt(l/g) where l is the length of the pendulum and g is the acceleration of gravity and as an approximation this does not depend on theta(0) the inital angle.