Simple pendulum oscillation

  • Thread starter Dx
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  • #1
Dx
Hello,

I have two test questions that I missed AT school so i wrote them down. I wanted to know the correct answers as well as a short description of how they came up with that.

1) a simple pendulum consists of a mass attached to a weightless string L. For this system when undergoing small oscillation

a. frequency proportional to amplitude -->that was my original answer
b. period proportional to amplitude
c. frequency independent of mass
frequency is independent of length

My new answer from what I read and understand would be d, am i correct?

2) Simple pendulum A swings back n forth at twice the frequency of a simple pendulum B. what statement is correct?

a. B is twice as long as A
b. B is twice as massive as A
c the length of B is 4 times the length of A
d. the mass of B is 4 times the mass of A

i dont remember but i think the formula is 1/2pi * sqrt(g/L)
anyways can anyone tell me the correct answer and briefly why?

Thanks!
Dx
 

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
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Dx, just look up the formulas for the frequency of a physical pendulum. That tells the whole story for #1. As for #2, you got the formula right, so use it to get the answer.

fA=(1/2π)(g/LA)1/2
fB=(1/2π)(g/LB)1/2

You are told that fA=2fB

Can you take it from there?
 
  • #3
Dx
Okay!

Ill take it tom. ill let you know how it went, k
Thanks!
dx :wink:
 

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