# Simple Harmonic Motion of a Pendulum

1. Homework Statement
What is the amplitude of a pendulum whose angle is given by (.10 rad)cos(5t + pi)
where t is in sec.

2. Homework Equations
s(t)=Acos(omega*t+phase constant)
Theta(t)=thetamax*cos(omega*t+phase constant)

3. The Attempt at a Solution
I'm not quite sure if I did this right:

I found the length of the string to be .392 m ( by using omega= sqrt(g/L)
So, using the small angle approximation s= .392 * sin theta where theta= the intial angle, which I found to be .1 rad by looking at the equation

Then I used s(t) =Acos(omega*t + phase constant) with t=0
I also found omega= 5 and the phase constant is pi

.039 m = A cos(pi)
But my amplitude ends up negative. I've having a bit of trouble with positive and negative signs in SHM

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alphysicist
Homework Helper
Hi bcjochim07,

I think there are several things that are not quite right. The first and maybe the cause of the others is your identification of 0.1 radians as the initial angle. (The initial angle would be found by setting t=0 and solving for theta.)

So what would you say the angle 0.1 radians is? (It's actually mentioned in your formula for theta(t).)

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alphysicist
Homework Helper
Another issue that does not lead to a numerical error is that you used

s= r sin(theta) to find s=0.039 m

This gives the correct numerical value because theta is small (as it has to be for the pendulum to be in SHM), but s is actually the arc length along the pendulum's circular path so we would have

s = r theta

But the number you get looks right; now identify what that number represents (because of the particular angle that you used).

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Actually theta = (.10 rad)cos(5t + pi) is the expression for the pendulum with the variable the angle.
so 0.10 rad is the amplitude.

alphysicist
Homework Helper
Hi bcjochim07,

In my two posts I asked you for the identification of two values that you found. The angle you found was the angular amplitude (as SimonZ stated) and the other (0.039 m) is the displacment amplitude (along the arc).

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