# Time it takes a pendulum to reach a point

1. Nov 3, 2012

### DRC12

1. The problem statement, all variables and given/known data
A 150g mass on a 2.2m long string is pulled 6.6° to one side and released. How long does it take for the pendulum to reach 2.8° on the opposite side

2. Relevant equations
I don't know

3. The attempt at a solution
I don't know where to start. I was thinking using energy but we don't have the equation for harmonic motion energy only works for springs. The other way I thought was to use x=Acos(ωt), v=-ωAsin(ωt), and a=-ω2Acos(ωt) but I can't get the right answer.

2. Nov 3, 2012

3. Nov 3, 2012

### TSny

x=Acos(ωt) is a good choice. Can you relate all the symbols in this equation to the pendulum problem you need to solve? [EDIT: Whoops, I see stepped in after frogjg2003 already started. Sorry. And I should also have asked DRC12 to show some details of what he/she already tried.]

4. Nov 3, 2012

### DRC12

ω=(g/L)^.5=2.11
T=2π/ω=2.98
x=Acos(ωt)
2.8=6.6cos(2.11t)
2.8/6.6=.424=cos(2.11t)
arccos(.424)=2.11t
t=.539s; Time it takes for pendulum to go from 2.8° to 6.6°
Then the time it to go from 6.6° to 2.8° on the other side is 2.98-.539=2.44s

5. Nov 3, 2012

### TSny

Everything looks good except the use of the period T here.

6. Nov 3, 2012

### DRC12

but I'm looking for the time it takes for the pendulum to go from 6.6° to 2.8° on the opposite side not the same side

7. Nov 3, 2012

### TSny

Right. What's the defintion of "period"?

8. Nov 3, 2012

### frogjg2003

Remember, it's 2.8 degrees on the opposite side. What is the x position then?

9. Nov 3, 2012

### DRC12

The time it takes to go from A to -A and bak then the time should be T/2-.53=.96s

since it's on the other side it would be 9.4 but if you plug that into the equation 9.4=6.6cos(ωt)
arccos(1.42)=ωt but 1.42 is out of the range of cos

10. Nov 3, 2012

### TSny

Right. Good. But frogjg2003 is giving you hints for an easier way. Remember, θ is measured from the vertical position of the pendulum.

11. Nov 3, 2012

### DRC12

I'm still confused about what frogjg2003 is saying

12. Nov 3, 2012

### TSny

What is the value of x when the pendulum is 2.8 deg on the opposite side? It can't be 9.4 deg. That would make the pendulum higher than it started.

Last edited: Nov 3, 2012
13. Nov 3, 2012

### frogjg2003

Plug t=.96s into $x=A\cos(\omega t)$. What value do you get for x? What's so special about that value?
Note: There will be some rounding errors, ignore those.

14. Nov 3, 2012

### DRC12

It's negative got it. Thanks both of you