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Degrees, radians, angular velocity and phase angle

Posted in mathematics, Physics, tagged angular velocity, Babylonians, Degrees, Moon cycle, period, Radians, wave motion on 02/06/2013| 9 Comments »

Why are there $360^{\circ}$ in a circle?

I would imagine all of you reading this know that when we learn about angles, we learn that we measure angles in degrees, and that there are $360^{\circ}$ in a full circle. Why this strange number?

It was the Babylionians who chose to have $360^{\circ}$ in a full circle. When you think about it, they could have chosen any figure, say 400 or 10 or 1,000. Why did they choose 360? Well, it turns out that 360 was a special number to the Babylonians, or to be more correct 12 and 30 were, and 360 is 12 multiplied by 30.

12 was special to the Babylonians because there are 12 and a bit cycles of the Moon in a year. It is from the time between e.g. each new Moon (or full Moon) that we get our idea of months. 30 was special to the Babylonians because this was, roughly speaking, the number of days in each lunar cycle.

So, if 12 and 30 were special, then the number produced by multiplying them together, 360, was even more special. That’s why we are stuck with the strange number for the degrees in a circle.

Radians – more natural units

Mathematicians, physicists and a lot of other scientists most often use a different unit to measure angles, a radian. It is a more natural unit than degrees. A radian is very simply defined. If we have a circle with a radius $r$ , and we draw an angle $\theta$ as shown in the diagram below, then we will call the length of the arc produced $l$ .

The definition of the radian is quite simply that if $\theta$ is measured in radians then

$\boxed { \theta = \frac{l}{r} }$

Converting between degrees and radians

Converting between these two ways of measuring an angle is quite simple. Remember that the circumference of a circle, the distance all the way around, is $2\pi r$ . But this is just the length $l$ of our arc when $\theta = 360^{\circ}$ . So when $\theta=360^{\circ}$ this is equivalent to $\theta = \frac{2\pi r}{r} = 2 \pi$ radians.

This table shows the conversion from degrees to radians for some commonly used angles.

Degrees	Radians
0	0
30	$\frac{\pi}{6}$
45	$\frac{\pi}{4}$
60	$\frac{\pi}{3}$
90	$\frac{\pi}{2}$
180	$\pi$
270	$\frac{3\pi}{2}$
360	$2 \pi$

The sine function

The sine function, $\sin(\theta)$ , can be multiplied by a factor to produce a regularly varying function of any amplitude. For example, $4 \sin(\theta)$ would look like this (remember we are expressing $\theta$ in radians, not degrees).

The sine function is very useful in many areas of mathematics and physics.

Time varying waves

Suppose we have a wave travelling out from a stone which has been dropped into the water in a smooth pond. The ripples (waves) will travel outwards from where the stone was dropped. If we look at the height of the water at some point on the surface of the pond, then we find that the height $y$ of the water will vary with time $t$ as $y=A + R\sin (\omega t + \phi)$ . In this expression $A$ is the mean (average) height of the water (the level of the water before we dropped the stone), $R$ is the size of the wave (the peak height), $\omega$ is something called the angular velocity, $t$ is time (in seconds), and $\phi$ is something called the phase angle.

Angular velocity

Linear velocity, the velocity of something moving in a straight line in, say, the x-direction is just defined as $v=\frac{x}{t}$ . If an object is moving in a circle, we can define a similar concept called the angular velocity, which is not the linear distance divided by time, but rather the angular distance divided by time. But, the angular distance is just $\theta$ (when expressed in radians), so the angular velocity $\omega = \frac{ \theta }{t}$ and is measured in radians per second. Re-arranging this, we can see that $\theta = \omega t$ , so in other words $\omega t$ has the same units as radians, they multiply together to measure an angle. This means that $\sin{\theta}$ is equivalent to $\sin(\omega t)$ , enabling us to consider the variation of a wave as a function of time.

When expressed in radians, the angle for one complete cycle of a sine curve is just given by $\theta = 2\pi$ . Using our relationship between $\theta$ and $\omega t$ we can say that the time taken for one complete cycle, which we called the period $T$ is just given by $T = \frac{2\pi}{\omega}$ . So, depending on the value of the angular velocity $\omega$ , this determines how many seconds it takes for the sine function to complete one cycle.

The phase angle

The phase angle, $\phi$ above, is needed as our zero time may not be when the expression $R \sin( \omega t)$ is zero. So, it just allows us to correct for when we start timing, just in case this is not at the moment when our sine function has zero value.

Putting this together

Below are plots of two sine curves, the first (the blue curve) is given by $y = 4 \sin ( \frac{\pi}{4} t)$ . The second one (the green curve) is of $y = 4 \sin ( \frac{\pi}{4} t + 0.5 )$ (remember the phase angle $\phi$ is expressed in radians, and one radian is approximately $57^{\circ}$ .

Because $\omega=\frac{\pi}{4}$ , this means the period of the wave will be $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/4} = 8s$ . You can see that the blue curve does indeed complete one cycle in 8 seconds. The green curve has the same value of $\omega$ , so it has the same period. However, it is shifted in time due to the phase angle $\phi$ . So, at a time of $t=0$ , the green curve is not zero, and we say that it is ahead or leads the blue curve as it reaches its first peak before the blue curve does. The number of seconds it leads the blue curve can be found by considering that $\phi = \omega t$ so $t = \frac{\phi}{\omega}$ , which in this example is $t = \frac{1/2}{\pi/4} = \frac{2}{\pi} \approx 0.63 s$ .

In tomorrow’s blog I will discuss an exam question I have been going over with my students, where the height of water in metres in a harbour is given by e.g. $y = 6\sin(\frac{\pi}{4}t) + 8\cos(\frac{\pi}{4}t) + 11$ (where $t$ is expressed in hours). I will discuss how we can work out the maximum and minimum heights of the water, when e.g. the first maximum occurs, and how many hours a day we can use the harbour if we need a minimum of e.g. 2 metres of water in the harbour.

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