Simple harmonic motion and amplitude of an object

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SUMMARY

The discussion centers on calculating the time it takes for an object undergoing simple harmonic motion to travel from its maximum amplitude (A) to half of that amplitude (A/2). The established answer is T/6, derived from the equation x = A*cos(2π/T * t). By solving for t when x equals A and A/2, the time difference is determined, confirming that at x = A/2, t equals πT/6.

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  • Understanding of simple harmonic motion principles
  • Familiarity with trigonometric functions, specifically cosine
  • Knowledge of the relationship between period (T) and amplitude (A)
  • Ability to manipulate and solve equations involving variables
NEXT STEPS
  • Study the derivation of the equations of motion for simple harmonic oscillators
  • Learn about the implications of amplitude and period in oscillatory systems
  • Explore the applications of simple harmonic motion in real-world scenarios
  • Investigate the effects of damping on simple harmonic motion
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Students studying physics, educators teaching mechanics, and anyone interested in the mathematical modeling of oscillatory systems.

wilmerena
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I have a short question:

an object undergoes simple harmonic motion with a period T and amplitude A. How long does it take the object to travel from x = A to x = A/2 ?

the answer is T/6, but I am not sure how to get to that,

Do I get it from x = Acos(2pi/T xt)?
help :cry:
 
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You could (but I wouldn't use an x to stand for multiplication especially in an equation that already has an x in it). You could solve for t when x = A and when x = A/2 and find the difference in times.

You know that x = A at t = 0.

for x = A/2:

[tex]x = \frac A 2 = A\cos \left( \frac{2\pi t}{T} \right )[/tex]

The cosine of 60 degrees is 1/2, so this reduces to:

[tex]\frac \pi 3 = \frac{2\pi t}{T}[/tex]

Solve for t and you'll find the answer.
 
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thanks so much :smile:
 

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