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## Homework Statement

A 180 g mass on a 2.5m long string is pulled 6.6 degrees to one side and released. How long does it take for the pendulum to reach 2.7 degrees on the opposite side?

## Homework Equations

omega = sqrt(g/L)

theta = S/L

theta(t) = A * cos(omega * t + phi)

## The Attempt at a Solution

I am looking at this problem in two parts, one is the time it takes the pendulum to swing 6.6 degrees (0.12 rad) and the other is how long it takes for the pendulum to go the remaining 2.7 degrees (0.047 rad). I have:

omega = sqrt(g/L) = 1.98 rad/s

For time it takes to swing 0.12 rad:

theta = S/L => s = L * theta => s = 0.3m = A

theta(t) = A * cos(omega * t + phi) => 0.12rad = 0.3 * cos(1.98t) => t = 0.59s

For time it takes to swing 0.047 rad:

theta = S/L => s = L * theta => s = 0.118m = A

theta(t) = A * cos(omega * t + phi) => 0.047rad = 0.118 * cos(1.98t) => t = 0.59s

Now, I know for a fact the second part is wrong as the pendulum doesn't go nearly as far. For the first part I left the phase constant as 0 since the pendulum is being held to the far right of its motion, however for the second part I'm not sure what to do with the phase constant. I know that the phase constant should probably be negative as it is pendulum motion to the left, however I'm not sure how to find it. I think if I can find the phase constant for the second part, use it to find the time, and add the times together I should get total time.

Any help is appreciated.