Why does time period not depend on amplitude?

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The Simple Pendulum
Pendulums of Arbitrary Shape
Variation of Period of a Pendulum with Amplitude

Michael Fowler

The Simple Pendulum

Galileo was the first to record that the period of a swinging lamp high in a cathedral was independent of the amplitude of the oscillations, at least for the small amplitudes he could observe. In 1657, Huygens constructed the first pendulum clock, a vast improvement in timekeeping over all previous techniques. So the pendulum was the first oscillator of real technological importance.

In fact, though, the pendulum is not quite a simple harmonic oscillator: the period does depend on the amplitude, but provided the angular amplitude is kept small, this is a small effect.

The weight mg of the bob (the mass at the end of the light rod) can be written in terms of components parallel and perpendicular to the rod. The component parallel to the rod balances the tension in the rod. The component perpendicular to the rod accelerates the bob,

ml d 2 θ d t 2 =−mgsinθ.

The mass cancels between the two sides, pendulums of different masses having the same length behave identically. (In fact, this was one of the first tests that inertial mass and gravitational mass are indeed equal: pendulums made of different materials, but the same length, had the same period.)

For small angles, the equation is close to that for a simple harmonic oscillator,

l d 2 θ d t 2 =−gθ,

with frequency ω= g/l , that is, time of one oscillation T=2π l/g . At a displacement of ten degrees, the simple harmonic approximation overestimates the restoring force by around one part in a thousand, and for smaller angles this error goes essentially as the square of the angle. So a pendulum clock designed to keep time with small oscillations of the pendulum will gain four seconds an hour or so if the pendulum is made to swing with a maximum angular displacement of ten degrees.

The potential energy of the pendulum relative to its rest position is just mgh, where h is the height difference, that is, mgl( 1−cosθ ). The total energy is therefore

E= 1 2 m ( l dθ dt ) 2 +mgl( 1−cosθ )≅ 1 2 m ( l dθ dt ) 2 + 1 2 mgl θ 2

for small angles.

Pendulums of Arbitrary Shape

The analysis of pendulum motion in terms of angular displacement works for any rigid body swinging back and forth about a horizontal axis under gravity. For example, consider a rigid rod.

The kinetic energy is given by 1 2 I θ ˙ 2 , where I is the moment of inertia of the body about the rod, the potential energy is mgl( 1−cosθ ) as before, but l is now the distance of the center of mass from the axis.

The equation of motion is that the rate of change of angular momentum equals the applied torque,

I θ ¨ =−mglsinθ ,

for small angles the period T=2π I/mgl , and for the simple pendulum we considered first I=m l 2 , giving the previous result.

Variation of Period of a Pendulum with Amplitude

As the amplitude of pendulum motion increases, the period lengthens, because the restoring force −mgsinθ increases more slowly than −mgθ ( sinθ≅θ− θ 3 /3! for small angles).

The simplest way to get some idea how this happens is to explore it with the accompanying spreadsheet.

Begin with an initial displacement of 0.1 radians (5.7 degrees):

Next, try one radian:

The change in period is a little less that 10%, not too dramatic considering the large amplitude of this swing.

Two radians gives an increase around 35%, and three radians amplitude increases the period almost threefold.

It’s well worth exploring further with the spreadsheet!

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A simple pendulum consists of a ball (point-mass) m hanging from a (massless) string of length L and fixed at a pivot point P. When displaced to an initial angle and released, the pendulum will swing back and forth with periodic motion. By applying Newton's secont law for rotational systems, the equation of motion for the pendulum may be obtained $$ \tau = I \alpha \qquad \Rightarrow \qquad -mg \sin\theta\; L = mL^2 \; \frac{d^2\theta}{dt^2} $$ and rearranged as $$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0 $$ If the amplitude of angular displacement is small enough, so the small angle approximation ($\sin\theta\approx\theta$) holds true, then the equation of motion reduces to the equation of simple harmonic motion $$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0 $$ The simple harmonic solution is $$\theta(t) = \theta_o \cos(\omega t) \ , $$ where $\theta_o$ is the initial angular displacement, and $\omega = \sqrt{g/L}$ the natural frequency of the motion. The period of this sytem (time for one oscillation) is $$ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}} . $$

The period of a pendulum does not depend on the mass of the ball, but only on the length of the string. Two pendula with different masses but the same length will have the same period. Two pendula with different lengths will different periods; the pendulum with the longer string will have the longer period.

How many complete oscillations do the blue and brown pendula complete in the time for one complete oscillation of the longer (black) pendulum?

From this information and the definition of the period for a simple pendulum, what is the ratio of lengths for the three pendula?

With the assumption of small angles, the frequency and period of the pendulum are independent of the initial angular displacement amplitude. A pendulum will have the same period regardless of its initial angle. This simple approximation is illustrated in the animation at left. All three pendulums cycle through one complete oscillation in the same amount of time, regardless of the initial angle.

When the angular displacement amplitude of the pendulum is large enough that the small angle approximation no longer holds, then the equation of motion must remain in its nonlinear form $$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0 $$ This differential equation does not have a closed form solution, but instead must be solved numerically using a computer. Mathematica numerically solves this differential equation very easily with the built in function NDSolve[ ].

The small angle approximation is valid for initial angular displacements of about 20° or less. If the initial angle is smaller than this amount, then the simple harmonic approximation is sufficient. But, if the angle is larger, then the differences between the small angle approximation and the exact solution quickly become apparent.

In the animation below left, the initial angle is small. The dark blue pendulum is the small angle approximation, and the light blue pendulum (initially hidden behind) is the exact solution. For a small initial angle, it takes a rather large number of oscillations before the difference between the small angle approximation (dark blue) and the exact solution (light blue) begin to noticeable diverge.

In the animation below right, the initial angle is large. The black pendulum is the small angle approximation, and the lighter gray pendulum (initially hidden behind) is the exact solution. For a large initial angle, the difference between the small angle approximation (black) and the exact solution (light gray) becomes apparent almost immediately.

Hand-wavy intuition: Suppose we don't know about pendulums but want to construct a one-dimensional path, such that a point mass constrained to this path can oscillate around a low point with different amplitudes but constant period.

We do this from the bottom up -- so imagine that we have constructed the path from an altitude of $h_1$ down to $0$ and back up to $h_1$ on the other side. Now we want to extend that up to a sligher higher altitude $h_2$.

When we release our point mass at $h_2$, we can compute what its kinetic energy (and therefore its speed) will be at any altitude, so we can (at least in principle) compute how long it takes it to pass by the already made $h_1$-to-$h_1$ segment. This will be less time than our desired period, and half of the remaining time will be how long the point mass should take to move from $h_2$ to $h_1$. If that is sufficient time (that is, more than it would take the mass to fall straight down from $h_2$ to $h_1$), we can adjust how long time it takes, simply by making the section from $h_2$ to $h_1$ a suitably inclined plane.

Taking this process to the limit (where $h_2$ is just infinitesimally higher than $h_1$) we get some horrendous continuous-delay differential equation that I don't care to either derive in detail or solve -- but Huygens did it in 1659 and found that the solution is an inverted cycloid.

So if we have a bob gliding frictionlessly along a cycloid, it will indeed have the same period for any amplitude.

A pendulum, of course, swings in a circle rather than a cycloid -- but the cycloid turns out to be smooth enough (with nonzero but finite curvature) at the bottom that it can be approximated by a circle. This approximation is good enough that the difference in period between the circle and cycloid go to $0$ as the amplitude goes to $0$.