Solving the Synchronized Oscillation of Two Pendulums

[idkgirl]

Homework Statement

Consider the two “gigantic” simple pendulums with identical masses but with different lengths
as shown below. Suppose they are released from rest from position A at the same time as
shown. So you understand that they will not oscillate in harmony since they will have different
periods of oscillation. But at a later time we will see both pendulums reaching the position A
simultaneously. When will it take place? (calculate the time taken from the start). Take
g=980 cm/s2 . Must show all the calculations.

Homework Equations

T = 2pi * square root of length/gravity

The Attempt at a Solution

I really don't know what I am doing. I don't know how to relate two different periods. =(

 

[berkeman]

Homework Statement

Consider the two “gigantic” simple pendulums with identical masses but with different lengths
as shown below. Suppose they are released from rest from position A at the same time as
shown. So you understand that they will not oscillate in harmony since they will have different
periods of oscillation. But at a later time we will see both pendulums reaching the position A
simultaneously. When will it take place? (calculate the time taken from the start). Take
g=980 cm/s2 . Must show all the calculations.

Homework Equations

T = 2pi * square root of length/gravity

The Attempt at a Solution

I really don't know what I am doing. I don't know how to relate two different periods. =(

Welcome to the PF.

So what are the two different periods?

 

[idkgirl]

the periods are 1.8558 (the one with a length of 85.5) and 4.6398 (the one with a length of 534.4).

I think what I should do is 4.6398/1.8558 and then multiply the periods by a common multiplier, correct?

 

[berkeman]

the periods are 1.8558 (the one with a length of 85.5) and 4.6398 (the one with a length of 534.4).

I think what I should do is 4.6398/1.8558 and then multiply the periods by a common multiplier, correct?

You are on the right track -- you need to find the least common multiple to find when the are coincident in position again...

 

[idkgirl]

Oh cool! Thanks so much. I multiplied 1.8558 by 5 and the other number by 2 to get 9.28. I mean, I could multiply the one with 1.8558 by 2.5 to get 4.6398 secs, since by the time the longer pendulum swings the short one will have completed 2.5 cycles. but maybe the problem with that is that you can't have .5 of cycle ...so yeah. ...that is why I am multiplying it by whole numbers, right?