Simple Harmonic Motion

Introduction

As to the ratio of times of oscillation of bodies hanging from strings of different lengths, those times are as the square roots of the string lengths; or we should say that the lengths are as the doubled ratios, or squares, of the times. Thus if, for example, you want the time of oscillation of one pendulum to be double the time of another, the length of its string must be four times that of the other; of if in the time of one vibration of the first, another is to make three, then the string of the first will be nine times as long as that of the other. It follows from this that the lengths of the strings have to one another the inverse ratio of the squares of the numbers of vibrations made in a given time.

— Galileo Galilei, The Two New Sciences

Now we turn our attention to motion which occurs periodically -- repeated motion. We've already seen a bit of this concept with circular motion, and indeed, that is one place to start, since circular motion, pendulum motion, and wave motion can all be explained by the same kind of trigonometric analysis.

Harmonic Motion

There is also an excellent University of Guelph tutorial on simple harmonic motion, and tons of simulations on the web.

Now that we have the basic concepts of force, motion, acceleration, energy and momentum in place, let's look at a more complex situation: repeated motion. There are many examples of motion which repeats itself within a fixed time period: a pendulum swinging, a spring mass bouncing, a ball rolling around in the bottom of a bowl. The concepts we need in order to deal with such periodic motions, vibrations, or oscillations have applications to other situations where a pattern repeats itself; in particular, we can use mechanical periodic motion as a key to what happens in wave phenomena.

Any motion where force is a linear function of displacement (depending only on the first power of x) is an example of harmonic motion.

Spring motion as harmonic motion

We have already scene the equation

F = - kx

or Hooke's law, which is used to describe the motion of a spring. Note that the force is is the opposite direction of the displacement—it always works to reduce x. Our k factor is a constant which incorporates the type of materials and such incidentals as temperature in the system; we will assume k remains constant for a given situation.

Because F = ma = -kx, where m and k are constants, a must vary as x varies. This means we cannot use the equations developed in chapter two, which depend on a being constant. We must analyze the situation using conservation of energy and momentum. To review, look at the diagram below. We ignore for the moment any potential energy due to position within the earth's gravitational field, and consider only the energy stored in the spring by forcibly extending it.

The total energy of the spring in motion is $E = KE + U = ½ m v^{2} + ½ k x^{2}$ (look back at chapter 6 if you need to check the derivation of this).

We consider four positions of special interest:

Before any displacement occurs, the system is at rest: $E = KE = U = 0$
As an initial condition, the spring has been extended to distance x = A. All energy is potential energy stored in the position of the spring, so $E = U = ½ k A^{2}$
Since this is the only source of energy in the system, the total energy is always proportional to the square of the maximum displacement, or amplitude.
When the spring is released, it has energy of position as well as motion, but the total energy must be equal to the initial potential energy stored in the extended spring: $E = KE + U = ½ m v^{2} + ½ k x^{2} = U_{\max} = ½ k A^{2}$
This brings us to the interesting conclusion that when the mass is passing through x = 0, all its energy is kinetic energy, and E = KE = mv²/2 . This velocity must be the maximum velocity that the system reaches. Moreover, the KE at this point must be equal to the original PE, so $E_{x = 0} = KE = ½ m v^{2}$
If we designate this maximum velocity as v₀, then we can reduce the relationship to $m {v_{0}}^{2} = k A^{2}$
This will come in useful later.

Now we can do some interesting substitutions, and solve for different situations, like the velocity as a function of displacement:

\begin{array}{l} ½ m v^{2} + ½ k x^{2} = ½ k A^{2} \\ m v^{2} + k x^{2} = k A^{2} \\ m v^{2} = - k x^{2} + k A^{2} = k {(A}^{2} - x^{2}) \\ v^{2} = \frac{k}{m} {(A}^{2} - x^{2}) \\ v = \sqrt{\frac{k}{m} {(A}^{2} - x^{2})} \end{array}

Circular motion as harmonic motion

Not surprisingly, we can make use of the fact that the regular motion of an object under constant centripetal force is also periodic motion. Observe the motion of the block on the spring and the block revolving around the circle at constant angular velocity = A * v₀. (Reload to force animation to run again).

If we use this analogy, then the period T for the repeated motion is the distance divided by velocity: $T = \frac{2 π A}{v_{0}}$ From the relationship above, however, we know that $m {v_{0}}^{2} = k A^{2} \Rightarrow \frac{m}{k} = \frac{A^{2}}{{v_{0}}^{2}} \Rightarrow \frac{A}{v_{0}} = \sqrt{\frac{m}{k}}$ so we can express the period (seconds per cycle) as $T = \frac{2 π A}{v_{0}} = 2 π \sqrt{\frac{m}{k}}$ and the frequency (cycles/second = Hertz) as $f = \frac{1}{T} = \frac{1}{2 π} \sqrt{\frac{k}{m}}$ Note the important result that the period (or frequency) does not depend on the amplitude!

We can now relate k and m back to angular velocity ω, which is simply distance in radians/time, and for a single full cycle, becomes $ω = \frac{2 π}{T} = \sqrt{\frac{k}{m}}$

Now we can do some important manipulations. Consider our analogy circle below:

Given the relationships we have already established for θ, we can express displacement variously as

\begin{array}{l} x = A \cos θ \\ x = A \cos ω t \\ x = A \cos (2 π ft) \\ x = A \cos (\frac{2 π t}{T}) \end{array}

But let's look a little more closely at the situation. We assume x = A where θ= 0, since cos 0 = 1. This means our initial conditions require that the spring be already extended. But that isn't always the case: we can have initial conditions where x = 0 if v > 0. So we need the more general condition

x = A \sin (ω t + δ)

where δ is a phase angle to be adjusted as needed for starting conditions. A brief look at the trigonometry of adding angles will make this clearer. We will use the trigonometric identity: $\sin (A + B) = \sin A \cos B + \cos A \sin B$ $x = A \sin (ω t + δ) = A [\sin ω t * \cos δ + \cos ω t * \sin δ]$

If we have a phase shift of δ = π/2 = 90º, then

x = A \sin (ω t + \frac{π}{2}) = A [\sin ω t + \cos 90 ° + \cos ω t + \sin 90 °]

Since cos 90º = 0 and sin 90º = 1, this reduces to

\begin{array}{l} x = A \sin (ω t + \frac{π}{2}) = A [\sin ω t * 0 + \cos ω t * 1] \\ x = A \cos ω t \end{array}

or our situation where x > 0 at t = 0. By manipulating δ, we can get the starting conditions we need for any situation.

Now let's go back to our earlier expression for velocity:

v = \pm \sqrt{\frac{k}{m} {(A}^{2} - x^{2})}

If we substitute in the general expression (assume for the moment that δ = 0) we have

\begin{array}{l} v = \pm \sqrt{\frac{k}{m} {(A}^{2} - {(A \cos ω t)}^{2})} \\ v = \pm \sqrt{\frac{k}{m} A^{2} (1 - \cos^{2} ω t)} \end{array}

Recall from trig that $\sin^{2} ω t + \cos^{2} ω t = 1 \Rightarrow 1 - \sin^{2} ω t = \cos^{2} ω t$ We can substitute $v = \pm \sqrt{\frac{k}{m} A^{2} (1 - \sin^{2} ω t)} \Rightarrow \pm \sqrt{\frac{k}{m} A^{2} (\cos^{2} ω t)} = \pm \sqrt{\frac{k}{m}} (A \cos ω t)$

Now we pull the rabbit out of the hat. Remember way up the page where we concluded that ω = √(k/m)? We can now substitute angular velocity in so that

v = ω * A \cos ω t = A ω \cos ω t

Finally, for acceleration, we derive

a = \frac{F}{m} = \frac{- kx}{m} = - \frac{k}{m} A \sin ω t = - ω^{2} A \sin ω t

Comparison of our results shows that velocity is 90° or π/2 out of phase with displacement, and acceleration is 180º out of phase with displacement. Velocity is therefore greatest when displacement is least, and acceleration is greatest when displacement is greatest, but acts in the opposite direction as the displacement—which was what we determined originally from F = -kx. Now, however, we have a general set of equations which will work for all cases of periodic motion.

Pendulum motion

Please review the derivation of pendulum motion in your text. The most important points are

$F = - mg \sin θ$ which results from a free body analysis of the gravitational force acting on the pendulum at any angle θ, and
$\sin θ \sim θ with 1 % for θ < 15 °$ This approximation allows us to reduce the force equation to F = - mg θ, which makes it equivalent in form to F =-kx, and therefore an example of harmonic motion.
From rotational motion,θ= x/L where x is the linear distance traveled and L is the radius of the circle, or in this case, the length of the pendulum. $F = - (mg / L) * x$ where mg/L is analogous to k.
From analysis of the period of the spring, we determined $T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{m}{\frac{mg}{L}}} = 2 π \sqrt{\frac{L}{g}}$

Practice with the Concepts

Discussion Points

Is the acceleration of a simple harmonic oscillator ever zero? If so, where?
Real springs have mass. Will the true. Infrequency be larger or smaller than given by the equations for a mass oscillating on the end of an idealized massless spring?

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Physics

Chapter 11: 1-6