Solving for T in a Simple Pendulum: Troubleshooting and Tips

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The discussion focuses on calculating the number of oscillations of a simple pendulum with a mass of 0.14 kg and a string length of 0.66 meters. The user derived the period (T) as 1.629 seconds for one oscillation using the formula L = (gT^2) / (4pi^2). To find the number of oscillations per minute, they divided 60 by the period, resulting in approximately 36.8 oscillations per minute. Other participants confirmed the calculation and noted that for a 1-meter length, the period is roughly 2 seconds. The user successfully solved the problem and received validation from peers.
fishingaddictr
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still reviewing my test and I am having trouble.

A pendulum consists of a bob of mass 0.14kg at the end of a light string of length 0.66 meters, whose other end is fixed. the number of oscillations executed by the pendulum each minute is...?

so far this is what I've got

L= (gT^2) / (4pi^2) solve for T, i get 1.629 seconds..

where do i go from here?
 
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i think i got it.. it takes 1.629 seconds for one oscillation.. we're solving for number of oscillations per min. so divide 60 by 1.629.. would be 36.8 oscillations per minute?

correct? anyone?
 
fishingaddictr said:
i think i got it.. it takes 1.629 seconds for one oscillation.. we're solving for number of oscillations per min. so divide 60 by 1.629.. would be 36.8 oscillations per minute?

correct? anyone?
Looks like you got it. A useful bit of trivia.. when L = 1 meter the period is very close to 2 seconds.
 

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