traitorman

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The question is from Periodic Motion chapter, passage 4, question 24.

It states "suppose the mass is pulled a small distance to the right of the equilibrium position and then released when t = 0. Which graph best represents velocity vs time?

The answers all include a sinusoidal wave on an x, y axis where v is on the y axis and t is on the x axis.

Based on the question, I assumed that at time = 0, the mass is at max displacement to the right so the graph should be the following:
 

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loveoforganic

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You have three main positions to consider.

1) Max displacement from equilibrium above x axis
2) Equilibrium position
3) Max displacement from equilibrium below x axis

Consider the potential and kinetic energy of the system at the three positions.

At 1, all of the energy in the spring is potential energy. There can be no velocity, because the spring has to reverse directions at some point, which necessitates a reversal in velocity. Acceleration is at the maximum.

At 2, all of the energy in the spring is kinetic energy. All of the potential energy from stretching the spring has been converted into returning the spring to its equilibrium position. However, now that it is in its equilibrium position, it has tons of velocity. Acceleration is at the minimum.

At 3, this velocity and kinetic energy is dissipated as the situation in 1 is mirrored. Acceleration is at the maximum.


At t=0, you're essentially at position 1 at the instantaneous moment of release.
 

traitorman

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ahhh im a fool. i agree with all your points but i realized the reason i still wasnt getting it was because its a velocity vs time graph not a position vs time graph.
 

Vanguard23

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1)
View attachment 14433


2)
View attachment 14434


Right is above the x axis and left is below the x axis. The answer is the second one. Why?? Their explanation is: At t = 0, the mass is at its max displacement and is not moving. This rules out choice 1 above.
Right is positive. Left is negative. If it's pulled to the right, then the spring pulls left(to the negative). Since this vector is a negative velocity, then the graph should correlate to an initial negative direction. Hence, it starts going down first.