Two Dimensional Harmonic Motion
This page is currently under construction. Further information will be added later, the example is present due to reorganization moving it away from the page Iterative Prediction of the Spring Mass System
Examples
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.
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