Simple harmonic motion and amplitude of an object

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An object in simple harmonic motion with a period T and amplitude A takes T/6 to travel from x = A to x = A/2. To find this, one can use the equation x = A*cos(2πt/T) and solve for t at both positions. At x = A, t is 0, and at x = A/2, the equation simplifies to cos(60 degrees) = 1/2, leading to t = T/6. This method effectively calculates the time difference required for the motion. Understanding these calculations is essential for analyzing simple harmonic motion.
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:

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

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

\frac \pi 3 = \frac{2\pi t}{T}

Solve for t and you'll find the answer.
 
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Likes Matthewp3247
thanks so much :smile:
 
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