Simple Harmonic motion of a Pendulum

[dolvo]

Homework Statement

A simple pendulum is made up of a 400-g mass attached to a 1.30-m string. It is pulled to the side and then released. Later a clock is started at the instant the mass is moving at its maximum velocity of 30 cm/s [R]. At what time t is the velocity of the mass 20 cm/s [L] for the third time?

Relevant Equations

w = sqrt(g/l)
v = xmax * w * cos(wt)

First, I decided to solve for the coefficient in front of the cosine simple harmonic function for velocity. I know there is max velocity of 30cm/s at time = 0 , so I plug it into velocity function.

xmax * w = A

v(t) = Acos(wt)
0.3 = Acos(w*0)
A = 0.3

Then I have my velocity function

v(t) = 0.3cos(wt)

So I now plug in -0.2 for velocity because that's what the question asks.

-0.2 = 0.3cos(wt)
-0.67 = cos(wt)

I then take inverse cosine , and get 2.3, in which I add 2pi to result in the third time the pendulum has the velocity. Then I divide by angular velocity

(w= sqrt(9.8/1.5) = 2.56)

2.3 + 2pi = wt
8.58 = 2.56t
t= 3.36

I get 3.36 s but the answer is 3.13!

 

[DrClaude]

Problem Statement: A simple pendulum is made up of a 400-g mass attached to a 1.30-m string.

(w= sqrt(9.8/1.5) = 2.56)