Simple harmonic motion and oscillation period

AI Thread Summary
An object in simple harmonic motion (SHM) oscillates with a period of 4.0 seconds and an amplitude of 10 cm, moving from x = 0.0 cm to x = 6.0 cm. The equation of motion is derived as x(t) = 0.10sin((π/2)t), utilizing the relationship between sine and cosine functions. A trigonometric identity shows that Acos(ωt - π/2) equals Asin(ωt), explaining the transformation. Two potential solutions arise from solving the equation, yielding times of 1.59 seconds and 0.41 seconds, with 0.41 seconds being the correct answer for the specified motion. The confusion stems from the interpretation of the cosine inverse function, where the negative solution is valid in this context.
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Homework Statement



An object in SHM oscillates with a period of 4.0 s and an amplitude of 10 cm. How long does the object to move from x = 0.0 cm to x = 6.0 cm?

Homework Equations



x(t) = Acos(ωt +φ )

The Attempt at a Solution



ω = π/2
Acosφ = 0 ⇒φ = ±π/2
Since object is moving to the right choose φ = -π/2

x(t) = Acos(ωt-π/2)
=Asinωt
=0.10sin(1/2)πt

My question is how why does Acos(ωt-π/2) = =Asinωt
 
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Acos(ωt-π/2) = =Asinωt

This is a standard trigonometric identity. The graph of the cosine function is a sine graph shifted by \pi/2. Thus, if you add or subtract \pi/2 from the argument of a sine function, you'll get the cosine function and vice versa.
 
OK, what if i solve this equation without using the trig idenitity

Apparently, there are then two possible answers,

either 0.927 = (pi/2)t - pi/2, which gives 1.59 s
or - 0.927 = (pi/2)t - pi/2. which gives 0.41 s

I know the answer is 0.41 seconds but why. Shouldnt I use the first equation because cos-1(0.06/0.01) is equal to postive 0.927
 
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