In this part you will be looking at how Newton's Second Law relates to Simple Harmonic Motion and the ideal spring. The spring is pulling to the left on the cart and gravity is pulling down on the hanging mass. Because of these forces, the cart accelerates back and forth, and the mass accelerates up and down. This time, the force sensor is zeroed at the cart's equilibrium position. This means that the force sensor is not reading the full force of the spring. It is reading how much more or less than the spring force at the cart's equilibrium position.

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Part C- Newton's Second Law and the Simple Harmonic Oscillator
In this part you will be looking at how Newton's Second Law relates to Simple Harmonic Motion and the
ideal spring. The spring is pulling to the left on the cart and gravity is pulling down on the hanging mass.
Because of these forces, the cart accelerates back and forth, and the mass accelerates up and down.
This time, the force sensor is zeroed at the cart's equilibrium position. This means that the force sensor
is not reading the full force of the spring. It is reading how much more or less than the spring force at
the cart's equilibrium position.
k
m.
force
sensor at equilibrium)
(reads zoro
position of mass
when spring is
position of cart Yo equilibrium
unstretched
Im,
- Yo
when spring is
unstretched
position
of cart
current
position
of cart
current
position
of mass
equilibrium
position
of mass
19. Analyze this as a physics 4A style force problem. Name the distance the spring is stretched, y.
Name the mass of the cart, m, and the mass of the hanger, m. Solve for the acceleration, a,
as a function of spring stretch. Remember the cart measures the positive direction to the left.
In other words, a positive acceleration is to the left.
Note that when the system is at rest in the equilibrium position, the spring is already stretched.
Transcribed Image Text:Part C- Newton's Second Law and the Simple Harmonic Oscillator In this part you will be looking at how Newton's Second Law relates to Simple Harmonic Motion and the ideal spring. The spring is pulling to the left on the cart and gravity is pulling down on the hanging mass. Because of these forces, the cart accelerates back and forth, and the mass accelerates up and down. This time, the force sensor is zeroed at the cart's equilibrium position. This means that the force sensor is not reading the full force of the spring. It is reading how much more or less than the spring force at the cart's equilibrium position. k m. force sensor at equilibrium) (reads zoro position of mass when spring is position of cart Yo equilibrium unstretched Im, - Yo when spring is unstretched position of cart current position of cart current position of mass equilibrium position of mass 19. Analyze this as a physics 4A style force problem. Name the distance the spring is stretched, y. Name the mass of the cart, m, and the mass of the hanger, m. Solve for the acceleration, a, as a function of spring stretch. Remember the cart measures the positive direction to the left. In other words, a positive acceleration is to the left. Note that when the system is at rest in the equilibrium position, the spring is already stretched.
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In the kxy - m(sub h)g portion (after Fn), where did the x go and what is the g? 

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