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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Chapter13: Vibrations And Waves
Section: Chapter Questions
Problem 14P: An object-spring system moving with simple harmonic motion has an amplitude A. (a) What is the total...
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Note that when the system is at rest in the equilibrium position, the spring is already stretched.
 
 
 
 
 
 
 
 
I don't want question 19 from the attached photo answered. Just this one:
 
20. Write an expression for how far the spring is stretched when the cart is at rest in the equilibrium
position. Name this distance y0.
In the data run, the positions are measured from the rest (equilibrium) position of the cart. Name this
position
y1 . The distance that the spring is stretched is y=y0 - y1. The minus sign is because a
positive position for the cart decreases the spring stretch. (See the attached image.) Substitute your expression from question 20 into y=y0-y1.
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.
Immmm
m
force (reads zero
sensor at equilibrium)
Y₁
position of cart Yo equilibrium
when spring is
unstretched
position
of cart
current
position
of cart
position of mass
when spring is
unstretched
current Yo
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, m₁- 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. Immmm m force (reads zero sensor at equilibrium) Y₁ position of cart Yo equilibrium when spring is unstretched position of cart current position of cart position of mass when spring is unstretched current Yo 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, m₁- 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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