Simple Harmonic Motion (Pendulum)

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Homework Help Overview

The discussion revolves around the analysis of two pendula, both with a length of 1.00m, but differing in mass (0.050kg and 0.100kg). Participants are exploring the periods of oscillation under various conditions, including small angular displacements and a larger angle of 60.0°.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the ratio of the periods of oscillation and question the role of mass in the period formula. There is an inquiry into the derivation of the formula for the period of oscillation when considering larger angular displacements.

Discussion Status

Some participants have provided initial calculations and expressed uncertainty about the correctness of their solutions. Others have raised questions about the assumptions made regarding mass and the derivation of the formulas used, indicating a mix of understanding and confusion.

Contextual Notes

There is mention of a lab manual providing the formulas, which may influence the understanding of their derivation and application. The discussion reflects a need for clarity on the assumptions underlying the formulas used in the context of pendulum motion.

KMcFadden
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Homework Statement


Two pendula of length 1.00m are set in motion at the same time. One pendula has a bob of mass 0.050kg and the other has a mass of 0.100kg.

1. What is the ratio of the periods of oscillation?

2. What is the period of oscillation if the initial angular displacement is small?

3. What is the period of oscillation if the initial angular displacement is 60.0°? Calculate the series out to three terms.


Homework Equations



1. T=2π√(L/g)

2. T=2π√(L/g)[1+1^2/2^2 〖sin〗^2 (θ/2)+(1^2 3^2)/(2^2 4^2 ) 〖sin〗^4 (θ/2)+(1^2 3^2 5^2)/(2^2 4^2 6^2 ) 〖sin〗^6 (θ/2)]

The Attempt at a Solution



1. 1:2 or 2:1 depending on which you consider mass 1 or mass 2.

2. T=2π√(L/g)
T=2π√(1.00m/〖9.808m/s〗^2 )
T=2.01s

3. T=2π√(L/g)[1+1^2/2^2 〖sin〗^2 (θ/2)+(1^2 3^2)/(2^2 4^2 ) 〖sin〗^4 (θ/2)+(1^2 3^2 5^2)/(2^2 4^2 6^2 ) 〖sin〗^6 (θ/2)]

T=2π√(1.00m/〖9.808m/s〗^2 )[1+1^2/2^2 〖sin〗^2 (60/2)+(1^2 3^2)/(2^2 4^2 ) 〖sin〗^4 (60/2)+(1^2 3^2 5^2)/(2^2 4^2 6^2 ) 〖sin〗^6 (60/2)]

T=2.006[1+0.5000+0.0088+0.00153]

T=2.006[1.51033]

T=3.03s
 
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Are my solutions correct?
 
KMcFadden said:

Homework Statement


Two pendula of length 1.00m are set in motion at the same time. One pendula has a bob of mass 0.050kg and the other has a mass of 0.100kg.

1. What is the ratio of the periods of oscillation?

2. What is the period of oscillation if the initial angular displacement is small?

3. What is the period of oscillation if the initial angular displacement is 60.0°? Calculate the series out to three terms.


Homework Equations



1. T=2π√(L/g)

2. T=2π√(L/g)[1+1^2/2^2 〖sin〗^2 (θ/2)+(1^2 3^2)/(2^2 4^2 ) 〖sin〗^4 (θ/2)+(1^2 3^2 5^2)/(2^2 4^2 6^2 ) 〖sin〗^6 (θ/2)]

The Attempt at a Solution



1. 1:2 or 2:1 depending on which you consider mass 1 or mass 2.

Is m included in the formula for T?

2. T=2π√(L/g)
T=2π√(1.00m/〖9.808m/s〗^2 )
T=2.01s

3. T=2π√(L/g)[1+1^2/2^2 〖sin〗^2 (θ/2)+(1^2 3^2)/(2^2 4^2 ) 〖sin〗^4 (θ/2)+(1^2 3^2 5^2)/(2^2 4^2 6^2 ) 〖sin〗^6 (θ/2)]

How is this formula derived? Where did you get it?
 
You should revise your concepts
 
The formulas used were those provided by my lab manual.
 
OK, let's start with part 1. Your formula is correct. So where do you get the idea that T is a function of mass?

In part 2, I realize you were given that formula. I probably should not have commented at all, but what I had in mind is that you're being given a formula without its prior derivation, and no way could you have been shown its derivation. So, bottom line, never mind on part 2.
 

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