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Posts Tagged ‘Swings’

A simple pendulum and circular motion

Posted in Physics, Science, Uncategorized, tagged , , , on 24/10/2013| 3 Comments »

Legend has it that it was Galileo who first noticed that the period of a pendulum’s swing does not depend on how large the swing is, but only on the length of the pendulum. The swinging back and forth of a pendulum is an example of a very important type of motion which crops up in many places in Nature, so called Simple Harmonic Motion (SHM). In this blog I will derive the basic equations of SHM, and then go on and talk about the deep connection between SHM and circular motion.

A swinging pendulum

If we start off by looking at a simple pendulum which has been displaced so that the bob is to the right of the vertical position, the angle the line of the pendulum makes with the vertical is given by \theta, and for this derivation to work \theta needs to be small.


A simple pendulum will swing back and forth, exhibiting Simple Harmonic Motion.

A simple pendulum will swing back and forth, exhibiting Simple Harmonic Motion.


The force restoring the pendulum bob back to the middle, which I have called F in the diagram above, is given by F = -T\sin(\theta) (the minus sign comes about because the the force is back towards the centre, even though the angle \theta increases as we move the bob to the right).

The restoring force, F, can be written using Newton’s 2nd law as F=ma=-T\sin(\theta). The angle \theta is measured in radians (see my blog here for a tutorial on radians). When \theta is small, \sin(\theta) \approx \theta and so we can write that



\sin(\theta) \approx \theta = \frac{x}{l}


where l is the length of the pendulum and x is the horizontal displacement of the pendulum bob.

Finally, the tension T in the pendulum chord can be written as T \approx mg where m is the mass of the bob and g is the acceleration due to gravity (9.8 m/s/s for the Earth).

Putting all of this together, we can write that the restoring force F can be written as


F = - mg\theta = -mg\frac{x}{l}

This means that the acceleration a can be written as


\vec{a} = -\frac{g}{l} \vec{x}

It is more common to write this as


\boxed{ \vec{a} = -\omega^{2} \vec{x} }


where \omega^{2} = \frac{g}{l} for the pendulum. \omega is called the angular frequency and it is related to how long the pendulum takes to complete one full swing, the period, by the equation T = \frac{2\pi}{\omega}. The frequency is just the reciprocal of the period, so we can write \nu = \frac{\omega}{2\pi} and so the angular frequency is related to the time frequency as \omega = 2\pi \nu.

Whenever the acceleration can be written as being proportional to the displacement, and in the opposite direction to the displacement, we have Simple Harmonic Motion. Other examples of SHM are an object bouncing vertically on a spring, or moving horizontally back and forth due to a spring attached at one end, even the vibrations of atoms in molecules.

Solutions to the SHM equation

What are the solutions to the second order differential equation \frac{d^{2}\vec{x}}{dt^{2}} = \vec{a} = - \omega^{2} \vec{x}? We have a displacement, \vec{x}, and it is proportional to the acceleration \vec{a}, but the acceleration acts in the opposite direction to the displacement.

We differentiate the displacement twice with respect to time to produce the acceleration (remember \vec{a} = \frac{d^2}{dx^2} \vec{x}), and for SHM, when we do this, the acceleration is proportional to the displacement and in the opposite direction.

Let us suppose we try the displacement



x = A \sin(\omega t) \text{ (remember that } \theta = \omega t)

If we differentiate this once with respect to time we get the velocity



v = \frac{dx}{dt} = \frac{d}{dt} A \sin(\omega t) = A\omega \cos(\omega t)

To get the acceleration we need to differentiate the velocity with respect to time so



a = \frac{dv}{dt} = \frac{d}{dt} A\omega \cos(\omega t) = - A \omega^{2} \sin(\omega t) = -\omega^{2}x

Loh and behold, we now have that a \propto - x, so we have shown that, if x = A \sin(\omega t) that an object which has this displacement as a function of time will display SHM.

As the acceleration is proportional to the displacement, it will be at its maximum when the displacement is maximum, so for a pendulum when the bob is at its extreme positions. The acceleration at the centre is zero, as the displacement at the centre is zero.

Conversely the velocity behaves in the opposite sense. Remember, as \vec{v} = \frac{d}{dt}\vec{x} = A\omega\cos(\omega t) it means that the velocity and displacement are 90^{\circ} out of phase with each other, when the displacement is a maximum the velocity is zero, and when the displacement is zero the velocity is a maximum. So, the bob will be travelling at its quickest when it passes through the centre, and at its extremes the velocity is (temporarily) zero.

SHM and circular motion

If an object is moving at a constant speed in a circle in the x-y plane we can write that it’s position at any time is given by



x = A \cos(\theta) \text{ and } y = A \sin(\theta)


For an object moving with a constant speed in a circle, its x-position and y-position can be written in terms of the radius A and the angular velocity.

For an object moving with a constant speed in a circle, its x-position and y-position can be written in terms of the radius A and the angular velocity \omega.


But notice that the expression for x is exactly the same as the expression which we had above, so an object moving in a circle performs SHM. But how? Surely, if it is moving with a constant speed (and hence constant angular velocity \omega), how is it also displaying SHM?

The SHM comes in when we look at the object’s x or y-position as a function of time. So, for example, if we look at the circle from below, as if we were looking along the plane of the screen from below, we would only see the x-displacement of the object, as the y-displacement would be invisible to us. The x-displacement is given by x=A \cos(\theta) = A \cos(\omega t), so the x-position will move back and forth about the central point, displaying SHM. This regular “wobble” is one of the things we look for in trying to find exoplanets – planets around other stars. If we see a regular, rhythmic wobble in the position of the parent star, it’s a pretty good bet that it has a planet going around it with the force of gravity between the host star and its planet producing the apparent SHM.


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