Two Dimensional Harmonic Motion

Main Idea

Mathematical Method

Two examples of harmonic motion in multiple dimensions will be considered: the pendulum and the swinging spring-mass system. In both cases, there exists the force of gravity pointing downwards, and another force pointed towards the center of rotation. We will not consider the analytic methods of solving these problems (for the first steps in doing so, see Centripetal Force and Curving Motion). Instead, we will define the forces present, and then implement an iterative model.

The Pendulum

The simpler of the two cases is the pendulum. In this scenario, we neglect air resistance, and have a mass attached to a rod of negligible mass with a fixed length [math]\displaystyle{ R }[/math]. We will treat the pivot point as the origin of the system, so that no coordinate conversions are necessary when computing angles. The force from the rod is a constraint force, meaning that it acts exactly to ensure the mass always stays on the arc of equal distance to the pivot point. For this to occur, it is necessary that the force directed towards the center always be the centripetal force necessary to maintain that arc, which is defined as a force pointed towards the pivot point with a magnitude

[math]\displaystyle{ |F_c| = \frac{v^2}{r} }[/math]

The situation is complicated by the fact that gravity is present. Gravity is defined as always:

[math]\displaystyle{ \vec{F}_g = -mg \hat{y} }[/math]

Now, let us imagine the situation, illustrated in this figure:

We release the pendulum from height [math]\displaystyle{ h }[/math] at an angle [math]\displaystyle{ \theta_0 }[/math]. At this point, its velocity will be zero, so the centripetal force must equal zero. However, gravity must still be present, so there must be some pendulum force to cancel gravity's radial component. From this we know its magnitude must be

[math]\displaystyle{ F_p (t) = mg\cos(\theta)+\frac{mv^2}{R} }[/math]

Now all we need to do is use trigonometry to determine the forces in the x and y direction:

[math]\displaystyle{ F_x = - F_p\sin(\theta) = -(mg\cos(\theta)\sin(\theta) + \frac{mv^2}{R}) }[/math]

Note the negative direction, because of how we have oriented our angles. Next, we have the slightly more complicated expression

[math]\displaystyle{ F_y = F_p\cos(\theta)-mg = mg\cos^2(\theta) + \frac{mv^2}{R}\cos(\theta) - mg }[/math]

Plugging these in to a numerical simulator will allow us to arrive at our desired answer.

Now, some interesting features of pendula may be noted. They will not be derived, but you are encouraged (see: the examples) to explore this behavior using a simulation. The most important feature can be noted by looking at the equations for force: each of the terms on the right has a mass term present, and on the left force is linearly dependent on mass. This means that the mass will actually fall out of the equations of motion: the motion will be determined solely by the initial conditions, the length of the pendulum, and the strength of gravity. You will see that the pendulum executes oscillatory motion, with an angular frequency

[math]\displaystyle{ \omega = \sqrt{\frac{g}{R}} }[/math]

Furthermore, for sufficiently small angles, this frequency is independent of the starting angle. The caveat about smaller angles is because, even in the absence of air resistance, is an approximation to say that the pendulum is a simple harmonic oscillator (specifically, it involves the approximation [math]\displaystyle{ \sin(\theta) = \theta }[/math]), but this approximation is very strong for smaller angles, so it's generally good enough.

The Swinging Spring

This problems is both more and less complicated than the one above. On the one hand, the force from the spring is not a centripetal restraining force, and so it will be more complicated to write out. On the other hand, it is conceptually easier to work with, because all we have to do is consider the two base forces and implement them, rather than thinking about the nature of the net force. First, gravity will stay the same as above, which should not be surprising. Next, we need to implement the spring force, as described in Simple Harmonic Motion and Iterative Prediction of Spring-Mass System. The Hookeian spring is governed by the equation

[math]\displaystyle{ \vec{F}_s = -k(\vec{r}-\vec{r}_0) }[/math]

This means that we define the rest length of the spring, then compute the current deviation of the spring from that rest length in order to determine the force. The force will always be directed along the axis of the spring, which in this cases means radially from the pivot point, as above (we will be assuming the spring stays straight - real springs will also have some curvature, but this can get very complicated). Given this, when the spring is at some angle [math]\displaystyle{ \theta }[/math], defined as above, we can determine our forces to be:

[math]\displaystyle{ \vec{F}_x =\vec{F}_s \cdot \hat{x} = k (\vec{r}-r_0\hat{r})\cdot\hat{x} }[/math]

This notation may be confusing, since it uses the inner product notation. This is meant to indicate that not only is the magnitude of [math]\displaystyle{ \vec{F}_s }[/math] important, but so is its sign. In the code, this is easy to implement, but notationally it is more difficult. One could explicitly write the equation out with trigonometric functions as

[math]\displaystyle{ \vec{F}_x = -|k (\vec{r}-r_0\hat{r})|(Sign(\vec{r}-r_0\hat{r}))\sin(\theta)\hat{x} }[/math]

but as you can see, it starts to get very clumsy, so instead this page will stick with vector notation. The coefficient sign is positive because we have one negative from the definition of a Hookeian spring, and another from our coordinate system (see above). Notice as well, the definition of the equilibrium point was made explicit, as we define it as being the point along the radial vector from the pivot to the object, with distance from the pivot of [math]\displaystyle{ r_0 }[/math]. The force in the y direction is slightly more complicated, but follow similarly:

[math]\displaystyle{ \vec{F}_y = \vec{F}_s\cdot\hat{y} -mg\hat{y} = -k(\vec{r}-r_0\hat{r})\cdot\hat{y} - mg\hat{y} }[/math]

Computational Method

The Pendulum

Given the forces we derived above, it is fairly straightforward to plug these into a numerical simulator along with a set of initial conditions. This does it using the numpy code described in Predicting Change in multiple dimensions.

The Swinging Spring

Given the forces we derived above, it is fairly straightforward to plug these into a numerical simulator along with a set of initial conditions. This does it using the numpy code described in Predicting Change in multiple dimensions.

Examples

Simple Example

Consider a pendulum consisting of a massless rod of length [math]\displaystyle{ 2.45 \; m }[/math] attached to an object of mass [math]\displaystyle{ 4 \; kg }[/math]. Assume that the acceleration due to gravity is equal to [math]\displaystyle{ 9.8 \; \frac{m}{s^2} }[/math], the initial angle is fairly small, and no air resistance is present. Compute the period of this pendulum. Confirm your result using a simulator

Solution

As long as we use the correct initial information and equations, this is a very straightforward problem. Remember that we have the formula for angular frequency

[math]\displaystyle{ \omega = \sqrt{\frac{g}{r}} }[/math]

Note that this formula does not include the mass of the object (although the assumption that the rod is important); the information about the mass was therefore a red herring. It is very straightforward then to compute that [math]\displaystyle{ \omega = 2 \; Hz }[/math]. Now, from this we can compute the period [math]\displaystyle{ T }[/math] using the relations

[math]\displaystyle{ \omega = 2\pi f }[/math]

[math]\displaystyle{ T = \frac{1}{f} }[/math]

where [math]\displaystyle{ f }[/math] is the normal frequency. Then we can substitute in to find that

[math]\displaystyle{ T = \frac{2 \pi}{\omega} = \frac{2 \pi }{2 \; Hz } = \pi \; s }[/math]

Now, let's confirm this with the simulator:

We can see that the period in x is indeed [math]\displaystyle{ \pi \; s }[/math]. The period in y is equal to [math]\displaystyle{ \frac{\pi}{2} \; s }[/math], which makes sense if one visualizes the pendulum: it will reach the same height both at its farthest left point and its farthest right point, so it will reach that height twice as often as it reaches its starting point. When we speak of the pendulum's period, however, we mean specifically the time it takes to return to its initial position, so the period in x is what we want.

Middling Example

Consider a massless spring of relaxed length 1 m with spring constant 120 N/m attached to the ceiling. If a 2 kg mass is released with an initial velocity of <1.5, 0.5, -0.5> m/s at an initial position of <-1, 1, 1> m, what is the velocity of the block after 0.005 seconds? Consider the point of attachment on the spring to the ceiling to be the origin. Only one iteration is necessary.

Solution

First, we calculate the initial momentum.

[math]\displaystyle{ {\vec{p}_{i}} = {{m}\cdot\vec{v}_{i}} }[/math]

[math]\displaystyle{ {\vec{p}_{i}} = {2kg\cdot{\lt 1.5, 0.5, -0.5\gt \; m/s}} = \lt 3, 1, -1\gt \;Ns }[/math]

As always, we have the force of gravity

[math]\displaystyle{ {\vec{F}_{grav}} = {m}{\lt 0, -g, 0\gt } }[/math]

[math]\displaystyle{ {\vec{F}_{grav}} = {2kg}\cdot{\lt 0, -9.8, 0\gt \; m/s^2} = {\lt 0, -19.6, 0\gt N} }[/math]

Now, a more complicated matter is the force of the spring

[math]\displaystyle{ \vec{F}_{spring} = -k(\vec{r}-\vec{r}_{0}) }[/math]

Using vector notation allows us to use this as a shortcut: Hooke's law is linear, so we can evaluate it separately for each components, as long as we get [math]\displaystyle{ \vec{r}_0 }[/math] right. To do this, we unit vector of the initial position:

[math]\displaystyle{ \hat{r} = \frac{\vec{r}}{|\vec{r}|} }[/math]

The magnitude of [math]\displaystyle{ r }[/math] is simply

[math]\displaystyle{ |\vec{r}| = \sqrt{(-1 \; m)^2 + (1 \; m)^2 + (1 \; m)^2} = \sqrt{3} \; m }[/math]

So then the unit vector will be

[math]\displaystyle{ \frac{\lt -1,1,1\gt \; m}{\sqrt{3} \; m} = \lt -\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}\gt }[/math]

Now, we have

[math]\displaystyle{ \vec{r}_0 = r_0 \hat{r} = (1\; m) \lt -\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}\gt = \lt -\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}\gt \; m }[/math]

And so the force from the spring will be

[math]\displaystyle{ '''Step 4: Caculate Net Force" The net force can be calculated by summing these two forces. \lt math\gt {\vec{F}_{net}} = {\vec{F}_{grav} +{\vec{F}_{spring}}} }[/math]

[math]\displaystyle{ {\vec{F}_{net}} = {\lt 0, -13.72, 0\gt N} +{\lt 12.83316, -29.94405, -8.55544\gt N} = {\lt 12.83316, -16.22405, -8.55544\gt N} }[/math]

Step 5: Update Momentum

Using the momentum update formula, calculate the momentum of the mass at the end of the given time step.

[math]\displaystyle{ {\vec{p}_{f} = \vec{p}_{i} + \vec{F}_{net}{&Delta;t}} }[/math]

[math]\displaystyle{ {\vec{p}_{f} = {\lt 1.68, 1.12, -0.28\gt Ns} + {\lt 12.83316, -16.22405, -8.55544\gt N}\cdot{0.005s} = { Ns}} }[/math]

[math]\displaystyle{ {\vec{p}_{f} = {\lt 1.7442, 1.0389 ,-0.3611\gt Ns}} }[/math]

Step 4: Calculate Velocity

[math]\displaystyle{ {\vec{v}_{f}} = \frac{\vec{p}_{f}}{m} }[/math]

[math]\displaystyle{ {\vec{v}_{f}} =\frac{\lt 1.7442, 1.0389 ,-0.3611\gt Ns}{1.4kg} = {\lt 1.2458, 0.7421, -0.2579\gt m/s} }[/math]

Step 5: State Answer

After 0.005 seconds, the mass is moving with velocity <1.25, 0.74, -0.26> m/s