We need to define a few terms related to harmonic motion. [b]The [i]period [/i]of the motion is the time it takes a cycle to repeat itself[/b]. The period is measured in seconds and is denoted by the letter T. [br][br]For instance, a pendulum's period is the time it takes to swing to and fro once. The object should be back where it started at the end of a period, and should be moving the same direction. Don't accidentally measure a half period by thinking that the time for a pendulum to go from its center, vertical orientation and back again is its period. If you start observing the pendulum while its moving leftward and passing center, it will go left, turn around, come through center again, go right, turn around and return to center a second time. That whole process takes one period.[br][br][b]The frequency [/b][math]f[/math][b] is the inverse of the period[/b], or [math]f\equiv \frac{1}{T}[/math]. So instead of seconds per oscillation, it is oscillations per second. In both cases the 'oscillations' do not have an associated unit. While the unit of frequency is inverse seconds, this unit is also called the hertz, or [math]1Hz\equiv 1s^{-1}.[/math][br][br]The angular frequency is closely related to the ordinary frequency except that it assumes that the motion goes through one cycle of [math]2\pi[/math] radians per period. This means that rather than speaking of cycles per second or oscillations per second, angular frequency is measured in radians per second. The lower case Greek omega is used to describe angular frequency, so we can write[math]\omega=2\pi f=\frac{2\pi}{T}.[/math] [br][br]While it may seem odd to speak of radians per second, you'll soon see that when we plot harmonic motion we see sine waves. Sine waves repeat every [math]2\pi[/math] radians, so it really is a sensible thing to do. Furthermore, when we study rotational motion we'll see the connection even more strongly between radians of angle and the motion, and all these definitions and the associated math will still apply. [br][br]What you do NOT want to do in the context of harmonic motion is somehow think about this angular frequency as relating to the actual motion of something like the arm of a pendulum and the rate at which it sweeps out angle. That is not what it is. Rather the angular frequency speaks only about the rate at which the periodic motion is repeated. In this sense its physical interpretation is more of a mathematical one.

When a mass is attached to a spring, pulled away from equilibrium and released, it leads to a vibration or oscillation.  The resulting vibrational motion is called [b]simple harmonic motion[/b], and is sometimes written as an acronym SHM.  The reason [i]simple[/i] is in the name is not that it's easy, but that in the event that the restoring force isn't linear (as is the case for a pendulum), or if there is damping (friction) involved, it becomes harmonic motion without the "simple" before it.  Harmonic just means repeating or rhythmic.[br][br][b]Simple harmonic motion will always result when a system is friction-free and subject to a linear restoring force[/b] as given by Hooke's law.  Mathematically, there are consequences associated with such a system:

[list=1][*]The amplitude of the motion will have no effect on the period of the motion. So a big oscillation of a certain mass on a certain spring will take just as long as a small oscillation.[/*][*]The position versus time will be perfectly described by a sinusoidal function. This means in terms of sine waves, cosine waves or linear combinations of the two.[/*][*]If the position is sinusoidal, then by definitions of velocity and acceleration from kinematics, so must the velocity and acceleration be sinusoidal since they are just time derivatives of positition.[br][/*][/list]

To work out the physics, all we need is Hooke's law and Newton's 2[sup]nd[/sup] law.  Putting them together gives:[br][br][center] [math]\sum F_x = -kx = ma.[/math][/center]In order to find the resulting motion we need to write this as a differential equation.  Recall from our studies of dynamics earlier that differential equations are just ones that contain both a function and one or several derivatives of the function.  Also recall that solving such equations means we must find a function that has a form that satisfies the equation.  Let's do this for the case of SHM.

In order to solve this equation we must first recognize that acceleration is the second derivative of position with respect to time.  Substituting this into the equation will give [math]-kx = m\frac{d^2x}{dt^2}.[/math]  The common thing to do with this is to combine terms on one side of the equation and leave a zero on the other side, and also to divide out the mass so that the second derivative is alone:[br][center][math]\frac{d^2x\left(t\right)}{dt^2}+\frac{k}{m}x\left(t\right)=0.[/math][/center][br]You need to recognize, as was generally the case in kinematics, that our position in the x-direction will depend on time. It is really x(t). You probably recognized that anyway since that was implied when rewriting the acceleration as the second derivative with respect to time, but I wanted to say it to be sure you recognized it.

To solve this equation we will, as we did in the chapter on variable forces, guess a trial function just by inspecting the equation.  We are guaranteed that if we guess a valid solution that it will be a unique and correct one[url=https://en.wikipedia.org/wiki/Ordinary_differential_equation#Existence_and_uniqueness_of_solutions]*[/url].  Guessing is not done blindly.  [br][br]Looking at the differential equation above, we see that in order to get a zero on the right side, we need the two terms to have opposite signs and identical functional form.  One of the terms includes the function itself, the other is the second derivative of the function. The only functions that have that behavior are sines, cosines and imaginary exponentials (which are closely related, but which we'll ignore for now).  So a general solution to the equation could take any of the following forms:[br][br][center][math]x(t)=A\cos(\omega t+\phi),[/math][/center] [br][center][math]x(t)=A\sin(\omega t+\phi),[/math][/center] [br][center][math]x(t)=A\cos(\omega t)+B\sin(\omega t).[/math][/center] [br]where[i] A[/i] and B are related to the amplitude of the motion, [math]\omega[/math] is the angular frequency, and [math]\phi[/math] is a phase shift to ‘move’ the functions left or right.  [br][br]Of course you didn't have to know ahead of time that we needed something called the angular frequency, but certainly it should be expected that some term needed to be next to the time since you can't take the cosine of time in seconds. Rather the arguments inside the trig functions need to be angles, and radians are what we always want. So the term multiplied by t had to have units of radians per second so that the argument [math]\omega t[/math] is in radians.[br][br][color=#1e84cc][i][u]A brief aside on angular units[/u][br]While it is common to use both degrees and radians for angles, in science and engineering, degrees are almost never used while doing the math. We use them to set up problems since people rather intuitively understand what a 30 degree angle or a 45 degree angle looks like. Once we begin doing math, however, you should use radians. It doesn't really matter when you are just looking for the sine of an angle, but when derivatives, integrals, arc lengths, frequencies and such are being used you need to use radians. The reason for this is that a radian isn't actually a unit. It's a ratio of the arc length to the radius of a circle! This is not true of degrees. When we use the arc length formula [math]s=r\theta[/math] we know to use radians. Really any time the math has any connection to circles, even in an abstract way, we need to use radians. This is one of those cases. So will be circular motion, orbital motion and all of the other wavy things in nature.[/i][/color][br][br]Just from the property of trig functions you should recognize that the larger the angular frequency, the quicker the oscillation or vibration. Recall that the relation between angular frequency and ordinary frequency (measured in cycles per second or Hz) is [math]\omega=2\pi f.[/math][br][br]

Since we have a general expression for the position of a simple harmonic oscillator given by [br][br][center][math]x(t)=A\cos(\omega t)+B\sin(\omega t),[/math][/center][br][br]we can calculate the corresponding velocity and acceleration of the oscillator. Taking one time derivative gives us the velocity along the x-direction: [br][br][center][math]v_x(t)=-\omega A\sin(\omega t)+\omega B\cos(\omega t).[/math][/center][br][br]Taking another derivative gives the acceleration:[br][br][center][math]a_x(t)=-\omega^2 A\cos(\omega t)-\omega^2 B\sin(\omega t).[/math][/center][br][br]Just looking at these expressions we can see that there are some phase relationships among the three kinematic variables of position, velocity and acceleration. Here is a plot with adjustable values for the parameters.

Knowing something about the initial conditions allows us to simplify the general expression we arrived at above. [b] Initial conditions are necessary to specify an exact solution[/b].  For instance, suppose that at t=0, the mass was at rest and the spring was stretched to its maximum amplitude A, and subsequently released.  [br]Knowing that allows us to set B=0 since otherwise the velocity dx(t)/dt would be non-zero at t=0, which can't be since the system started from rest. In mathematical terms, starting from rest implies:[br][br][center][math] v_x(t=0)=0 \\[br]v_x(t=0)=-\omega A\sin(0)+\omega B\cos(0)=\omega B \\[br]\therefore B=0. [/math][/center][br][br]That leaves us with [math]x(t)=A\cos(\omega t).[/math]  Furthermore, we can plug this solution into the differential equation and see that since [math]\frac{d^2x(t)}{dt^2}=-\omega^2Acos(\omega t),[/math] that the angular frequency must be related to the elastic constant and the mass by [math]\omega=\sqrt{\frac{k}{m}}.[/math]  To prove it simply requires you to take the second derivative of our x(t) solution with respect to time and plug it into the differential equation. It goes like this:[br][br][center][math]\frac{d^2x(t)}{dt^2} + \frac{k}{m}x(t) = 0 \\[br]-\omega^2Acos(\omega t)+\frac{k}{m}x(t)=0 \\[br]-\omega^2Acos(\omega t)+\frac{k}{m}A\cos(\omega t)=0 \\[br]\omega^2=\frac{k}{m} \\[br]\omega = \sqrt{\frac{k}{m}}. [/math][/center][br][br]If you look at the animation below, the connection between the motion of a mass on a spring and the sinusoidal solution we found above, should be clear.