Simple Pendulum Equations: Period, Length, and Angle as Functions of Time

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SUMMARY

The discussion focuses on the equations governing the motion of a simple pendulum, specifically its period and angular displacement as functions of time. The period of the pendulum is established as 2 seconds, leading to the conclusion that the length of the pendulum is approximately 1 meter, corresponding to option D. For the angular displacement, the correct equation is determined to be θ = θmax sin(πt), indicating that the angular frequency ω is π, derived from the relationship T = 2π/ω.

PREREQUISITES
  • Understanding of simple harmonic motion
  • Familiarity with the period of a pendulum formula
  • Knowledge of angular frequency and its relation to period
  • Basic grasp of Newton's second law of motion
NEXT STEPS
  • Study the derivation of the period of a simple pendulum using T = 2π√(l/g)
  • Learn about angular frequency in simple harmonic motion and its applications
  • Explore the relationship between force and motion in pendulum systems
  • Investigate the effects of varying gravitational acceleration on pendulum behavior
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Physics students, educators, and anyone interested in understanding the dynamics of simple harmonic motion and pendulum mechanics.

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A simple pendulum has a period of 2 s for small amplitude oscillations.

1. The length of the pendulum is most nearly
(A) l/6 m
(B) ¼ m
(C) ½ m
(D) 1 m
(E) 2 m

I used the period of a pendulum equation, did some algebra, and got D as the answer. Is this correct?


2. Which of the following equations could represent the angle θ that the pendulum makes with the vertical as a function of time t?

(A) θ = θmax sin (π/2)t

(B) θ = θmax sin πt

(C) θ = θmax sin 2πt

(D) θ = θmax sin 4πt

(E) θ = θmax sin 8πt

I don't get this one. I'm assuming it's C because 2π = ω in a lot of equations and maybe it's true here too? Can someone please explain this?
 
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1. Yes, good.

2. No. One way to think about this is, after one period the argument of the sine should be 2π. And you know what t is after one period.
 
I redid #2 this way.

T = 2π/ω
Substituted 2s for T and got ω = π

So the answer would be B?
 
Could someone help me out for #2. I'm not even sure how to approach that part.
 
bump, please help
 
An0maly said:
Could someone help me out for #2. I'm not even sure how to approach that part.

They solved it check the previous posts ..
 
I saw the answer but I didn't understand why they are plugging in T. What exactly is the concept of θ = θmax sin Wt? I've never learned that so I don't know how to manipulate the equation.
 
An0maly said:
I saw the answer but I didn't understand why they are plugging in T. What exactly is the concept of θ = θmax sin Wt? I've never learned that so I don't know how to manipulate the equation.

you can reach to the equation θ = θmax sin Wt by considering (i.e.) a pendulum of length l and start from Newton`s 2nd law which states that net force = m x`` , with little work you will end up with the solution θ = θmax sin Wt, where W = sqrt(g/l) ..

you can approach the same answer by substituting l = 1m, g=10m/s^2 in W , then you will get W = 3.16 for g = 10m/s^2 (or W=3.13 for g=9.8 m/s^2) which is approximately equal to the answer they reached .. :)
 

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