Finding Amplitude

  • #1

Homework Statement


An air-track glider attached to a spring oscillates between the 9.23 cm mark and the 50.23 cm mark on the track. The glider completes 27 oscillations in 44.7 s. angular frequency of the glider? (in rad/s)
3.7933
what is the amplitude of the glider?


Homework Equations



x=Acos(wt+ro)

The Attempt at a Solution


I considered that at x=9.23 the amplitude was zero and so is ro.
Hence, x=50.23 would equal just 50.23 cm=Acos(3.79(44.7)). But A turned out to be negative. What am I doing wrong? How do I solve this problem? Thank you in advance.
 

Answers and Replies

  • #2
dx
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2,014
18
What is the center of oscillation here? The amplitude is the maximum distance from the center that the glider reaches.
 
  • #3
Would it be half the distance the glider travels which is 20.5 cm?
 
  • #4
dx
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Correct.
 
  • #5
It still gives me a negative number if I do .205/((cos(3.79*44.7)).
 
  • #6
dx
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Negative number for what?
 
  • #7
For the amplitude A.
 
  • #8
dx
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2,014
18
:confused: You just said the amplitude was 20.5 cm, which is correct.
 
  • #9
OH! Ok I get it. Thank you very much.
 

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