How is the angular momentum related to x and y coordinates in SHM?

In summary, the content discussed is about two-dimensional simple harmonic motion and how the quantity x\dot{y}-y\dot{x} is constant along the ellipse. The second part of the conversation focuses on showing the relation between this quantity and the angular momentum of the system, using the vector definition of angular momentum for a particle.
  • #1
thatguy14
45
0

Homework Statement


Two-dimensional SHM: A particle undergoes simple harmonic motion in both the x and y directions
simultaneously. Its x and y coordinates are given by
x = asin(ωt)
y = bcos(ωt)

Show that the quantity x[itex]\dot{y}[/itex]-y[itex]\dot{x}[/itex] is also constant along the ellipse, where here the dot means the derivative with respect to time. Show that this quantity is related to the angular momentum of the system.

Homework Equations


L = mv x r

The Attempt at a Solution


Hi, so for the first part it is pretty simple and my answer is -abω, unless i made a dumb mistake which I don't think I did.

It's the second part that is giving me issues. How do I show that it is related to angular momentum? I tried doing this

L = [itex]\sqrt{L^{2}_{x}+L^{2}_{y}}[/itex]

then L[itex]_{x}[/itex] = m[itex]\frac{∂x}{∂t}[/itex] x r

where r = [itex]\sqrt{x^{2}+ y^{2}}[/itex]

and then plugging everything in. I was hoping all the cos and sin were going to cancel out but it got really huge and messy. I didn't think it was supposed to be that hard so can anyone tell me if I am going in the right direction or if there is something I am missing?

Thanks
 
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  • #2
Use the vector definition of angular momentum of a particle: ##\vec{L} = \vec{r} \times \vec{p}= m\; \vec{r} \times \vec{v}##, where ##\vec{r} = x\hat{i} + y\hat{j}##. (Note, the order of the cross product is important. Thus, ##\vec{L} = m\;\vec{v} \times \vec{r}## is not correct.)
 
  • #3
Right whoops.

so then v = dx/dt i + dy/dt j

and when we cross them we get m(xdy/dt - ydx/dt) = L correct?
 
  • #4
Yes. Good.
 
  • #5


The quantity x\dot{y}-y\dot{x} is known as the cross product of the position and velocity vectors, which is a measure of angular momentum. In this case, the position vector is (x,y) and the velocity vector is (dx/dt, dy/dt). So, the cross product can be written as:

x\dot{y}-y\dot{x} = (x,y) x (dx/dt, dy/dt)

= (x*dx/dt - y*dy/dt)

= (ax*cos(ωt) - by*sin(ωt)) - (ay*sin(ωt) + bx*cos(ωt))

= (a^2 + b^2)ωsin(ωt)cos(ωt) - (a^2 + b^2)ωsin(ωt)cos(ωt)

= 0

This shows that the quantity x\dot{y}-y\dot{x} is constant along the ellipse, as it is equal to zero at any point in time. This also means that the angular momentum, which is given by L = mv x r, is also constant along the ellipse. This makes sense because in SHM, the motion is periodic and the angular momentum is conserved.
 

What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object moves back and forth between two points along a straight line, with the same amount of time between each point. This type of motion occurs when the restoring force on the object is directly proportional to the displacement from its equilibrium position.

What are some examples of simple harmonic motion?

Some common examples of simple harmonic motion include a pendulum swinging back and forth, a mass on a spring bouncing up and down, and the motion of a tuning fork. These systems exhibit simple harmonic motion because they have an equilibrium position and a restoring force that follows Hooke's Law.

How is simple harmonic motion calculated?

The equation for simple harmonic motion is x = A * cos(ωt + φ), where x is the displacement from equilibrium, A is the amplitude (maximum displacement), ω is the angular frequency, and φ is the phase angle. The period of the motion can be calculated using the equation T = 2π/ω, and the frequency can be calculated using the equation f = 1/T.

What is the difference between simple harmonic motion and uniform circular motion?

While both simple harmonic motion and uniform circular motion involve periodic motion, they differ in their path and direction of motion. Simple harmonic motion occurs along a straight line, while uniform circular motion occurs along a circular path. Additionally, simple harmonic motion moves back and forth between two points, while uniform circular motion continues in the same direction.

What factors affect simple harmonic motion?

The factors that affect simple harmonic motion include the amplitude, mass of the object, and the restoring force. A larger amplitude will result in a longer period and a higher frequency. A greater mass will result in a longer period and a lower frequency. The strength of the restoring force also affects the motion, with a stronger force resulting in a shorter period and a higher frequency.

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