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

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Homework Help Overview

The discussion revolves around the equations governing a simple pendulum, specifically focusing on the period, length, and angle as functions of time. Participants are analyzing the relationship between these variables and attempting to determine the correct equations that describe the pendulum's motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the period of the pendulum and its relation to the length, with one suggesting a specific answer based on algebraic manipulation. Questions arise regarding the correct representation of the angle as a function of time, with some participants exploring the implications of angular frequency.

Discussion Status

Some participants express confidence in their answers, while others seek clarification on the reasoning behind the equations. There is an ongoing exploration of the relationship between the period and angular frequency, with multiple interpretations being considered.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available for discussion. There is a noted uncertainty regarding the manipulation of the equations and the underlying concepts of angular motion.

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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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