Time it takes a pendulum to reach a point

[DRC12]

Homework Statement

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

Homework Equations

I don't know

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.

 

[frogjg2003]

Could you show your work?

 

[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.]

 

[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

 

[TSny]

2.98-.539=2.44s

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

 

[DRC12]

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

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

 

[TSny]

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

 

[frogjg2003]

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

 

[DRC12]

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

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

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

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

 

[TSny]

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

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

 

[DRC12]

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

I'm still confused about what frogjg2003 is saying

 

[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.

 

[frogjg2003]

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

 

[DRC12]

It's negative got it. Thanks both of you