Simple Harmonic Oscillation Problem

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Homework Help Overview

The problem involves determining the first time after t=0.00 s at which the velocity of an object in simple harmonic motion, described by the equation v(t)= -(4.04m/s)sin(21.0t + 1.00π), equals -0.149m/s.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve the equation by setting -0.149 equal to -4.04 times the sine function but encounters a negative time. They also consider using cosine to eliminate the phase constant.
  • Some participants suggest modifying the sine argument by 2π and using a graphing calculator for better insight.
  • Others question the implications of changing the phase constant from π to -π and its effect on the solution.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the sine function and its periodicity. Suggestions for graphical analysis and adjustments to the phase constant have been offered, indicating a productive exchange of ideas without a clear consensus yet.

Contextual Notes

Participants are navigating potential flaws in the original poster's logic and considering the implications of periodicity in the sine function. There is an acknowledgment of the cyclical nature of the problem, which may influence the solutions being explored.

GarrettB
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Homework Statement


The velocity of an object in simple harmonic motion is given by v(t)= -(4.04m/s)sin(21.0t + 1.00π), where t is in seconds. What is the first time after t=0.00 s at which the velocity is -0.149m/s?


Homework Equations


N/A

The Attempt at a Solution


I thought this was an easy question but there must be a flaw somewhere in my logic. I've tried -0.149=-4.04*sin(21t+pi). However, solving for t gives me a negative number. I've also tried doing using cos instead of sin to get rid of the phase constant, but still the wrong answer. It's given me aprrox. 0.07s. Any help would be appreciated
 
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A couple of suggestions.
If you change the argument of sin by 2Pi what happens.
You might also use a graphing calculator to plot the function to get some insight as to what is going on.
 
Nothing happens if you change it by 2Pi? Since that's one full cycle. But this is 1pi?
 
You probably found the sin^-1 of .149/4.04 and that gave you a negative t. Could you change the 1pi to -p1 and resolve?
 
Of course you can, because its a cycle. Appreciate the insight.
 

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