Oscillation Problem -- Ball mass on the end of a horizontal spring

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Homework Help Overview

The discussion revolves around a problem involving a 200 g ball attached to a spring oscillating on a frictionless table. Participants are tasked with determining the amplitude of oscillation given the spring constant and the ball's velocity at a specific position.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between time, amplitude, and the signs of the values obtained. There is an exploration of how different solutions for time can affect the amplitude calculation.

Discussion Status

Some participants have provided insights into the implications of sign choices in their calculations, while others have noted that certain steps may not be necessary for finding the amplitude. The conversation indicates a productive exploration of the problem without reaching a definitive consensus.

Contextual Notes

There is mention of multiple potential solutions for time, which raises questions about the assumptions made during the calculations. Additionally, the relevance of trigonometric identities in the context of the problem is noted.

osten
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Homework Statement


A 200 g ball attached to a spring with spring constant 2.40 N/m oscillates horizontally on a frictionless table. Its velocity is 20.0 cm/s when x=−5.00cm.
What is the amplitude of oscillation?

Homework Equations


f=√(k/m) /2π
x(t) = Acos(2πft)
v(t) = -2πfAsin(2πft)

The Attempt at a Solution


I found that t = 0.137s, and A turned out to be negative, so it was probably wrong.
 
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osten said:
I found that t = 0.137s, and A turned out to be negative, so it was probably wrong.
You do not show your working, but there should be multiple solutions for t. Some will produce a positive A and some negative. What do you conclude from that?
 
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Thanks! I ignored the signs and got the right answer.
 
osten said:
Thanks! I ignored the signs and got the right answer.
Good. Do you understand why the choice of solution for t can switch the sign?
By the way, you did not need to find t at all. You could have used sin2+cos2=1.
 

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