Two Dimensional Harmonic Motion: Difference between revisions

Revision as of 12:53, 10 July 2019

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:

- - Insert figure here ***

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.

Examples

Simple Example

Middling Example

A semi-challenging problem, such as the one below, will often require you to perform vector calculations. However, the same steps as the easy example above still apply. Pay careful attention to signs as you solve these types of problems.

The simplest example of a spring mass system is one that moves in only one-direction.

Consider a massless spring of relaxed length 0.5 m with spring constant 120 N/m attached to the ceiling. If a 1.4 kg mass is released with an initial velocity of <1.2, 0.8, -0.2> m/s at an initial position of <-0.3, 0.7, 0.2> 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

Step 1: Calculate Initial Momentum The initial momentum is simply the product of the mass and velocity of the object.

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

[math]\displaystyle{ {\vec{p}_{i}} = {{1.4kg}\cdot{\lt 1.2, 0.8, -0.2\gt m/s}} = {\lt 1.68, 1.12, -0.28\gt Ns} }[/math]

Step 2: Calculate Force of Gravity

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

[math]\displaystyle{ {\vec{F}_{grav}} = {1.4kg}\cdot{\lt 0, -9.8, 0\gt m/s/s} = {\lt 0, -13.72, 0\gt N} }[/math]

Step 3: Calculate Force of Spring

[math]\displaystyle{ {\vec{F}_{spring}} = {-k}{(\vec{L}_{mag}-\vec{L}_{0})} \cdot {\hat{L}} }[/math]

The magnitude of L is equal to the square root of the squares of it's individual components

[math]\displaystyle{ {\hat{L}} = {\sqrt{ {{L}_{x}}^{2} + {{L}_{y}}^{2} + {{L}_{z}}^{2} }} }[/math]

[math]\displaystyle{ {\hat{L}} = {\sqrt{ {-0.3}^{2} + {0.7}^{2} + {0.2}^{2} }} = {\sqrt{0.62}} }[/math]

The unit vector of L ([math]\displaystyle{ {\hat{L}} }[/math]) is equal to L divided by its magnitude

[math]\displaystyle{ {\lVert\vec{L}\rVert} = {\frac{\vec{L}}{{\lVert\vec{L}\rVert}}} }[/math]

[math]\displaystyle{ {\lVert\vec{L}\rVert} = {\frac{\lt -0.3, 0.7, 0.2\gt m }{\sqrt{0.62}}} }[/math]

Substituting these values into Hooke's Law, then gives the force exerted by the spring:

[math]\displaystyle{ {\vec{F}_{spring}} = {-k}{(\lVert\vec{L}\rVert-\vec{L}_{0})} \cdot {\hat{L}} }[/math]

[math]\displaystyle{ {\vec{F}_{spring}} = {-120 N/m}{(\sqrt{0.62} m - {0.5 m})} \cdot {\frac{\lt -0.3, 0.7, 0.2\gt m }{\sqrt{0.62}}} }[/math]

[math]\displaystyle{ {\vec{F}_{spring}} = {\lt 12.83316, -29.94405, -8.55544\gt N} }[/math]

Step 4: Caculate Net Force"

The net force can be calculated by summing these two forces.

[math]\displaystyle{ {\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}{Δt}} }[/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

@@ Line 18: / Line 18: @@
 ***Insert figure here ***
-We release the pendulum from height <math> h </math> at an angle <math> \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. The trick, however, is that we ''do not'' need to figure out the pendulum force: we only care about the net forces when writing our simulation. The pendulum force is strictly radial (that is, towards the pivot point), and the radial net force is ''always'' the centripetal force. Therefore, the only tangential force will be gravity's tangential component, which we can find out easily enough with a little trigonometry. To be precise:
+We release the pendulum from height <math> h </math> at an angle <math> \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> F_{g,tangential} = mg\sin(\theta) </math>
-These are our forces. We need them in Cartesian coordinates, however, so we do a little bit of trigonometry. Remembering that <math> F_g </math> is only in the y direction, we have first that
-<math> F_x = -|F_c|\sin(\theta) \hat{x} </math>
-Note that this is negative, since for positive theta the mass is being pulled in the negative x direction. Now, the y force is somewhat more complicated, since we need to decompose tangential and radial forces separately, rather than thinking about gravity and the pendulum force:
-<math> F_y = (|F_c|\cos(\theta) - mg\sin^2(\theta))\hat{y} </math>
-This decomposition is a little odd, so it may be worth stopping to think about. The component of the radial force should be intuitive, since it is the complement to the component used to find the x component. For the component of the tangential force, we have made two projections: first, of gravity into the tangential direction, then of the tangential force back into the y direction, hence the <math> \sin^2(\theta) </math>. To convince yourself, imagine the situation at <math> \theta = 0,90^\circ </math>. In the first case, we want the centripetal force to be the net force in the y direction, and the equation does indeed result in this. In the second case, we want gravity to be the only force in the y direction, and the equation once again fulfills this. Now, we'll need to see how this works in the computational model, which is discussed below.
+<math> 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> 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> 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.
 ==Examples==