# Simple harmonic motion and oscillation period

## 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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G01
Homework Helper
Gold Member
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