Simple Harmonic Motion (Pendulum)

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SUMMARY

The discussion focuses on the analysis of two pendula, each with a length of 1.00m, where one has a mass of 0.050kg and the other 0.100kg. The ratio of their periods of oscillation is determined to be 1:2, with the period for small angular displacements calculated using the formula T=2π√(L/g), yielding a period of approximately 2.01 seconds. For an initial angular displacement of 60.0°, the period is calculated using a series expansion, resulting in a period of approximately 3.03 seconds.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM)
  • Familiarity with the formula T=2π√(L/g)
  • Knowledge of trigonometric functions, specifically sine
  • Ability to perform series expansions for small angles
NEXT STEPS
  • Study the derivation of the formula T=2π√(L/g) for pendulums
  • Explore the effects of mass on the period of oscillation in different systems
  • Learn about the impact of angular displacement on the period of oscillation
  • Investigate advanced topics in oscillatory motion, such as damping and resonance
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to pendulum dynamics.

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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