Solve SHM Eqn Problem: Max Energy Transformation

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Homework Help Overview

The discussion revolves around a problem involving simple harmonic motion (SHM) where a mass on a spring oscillates. The position of the mass is described by the equation x = (4.55 cm) cos(3.70t rad/s). Participants are tasked with determining when the potential energy of the system transitions most rapidly into kinetic energy during the first cycle, as well as finding the maximum rate of energy transformation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to differentiate the potential energy function and question the correct approach to find the maximum rate of change of energy. Some express confusion about the relationship between potential and kinetic energy and how to derive the necessary equations. Others suggest focusing on the rate of change of potential energy rather than its maximum value.

Discussion Status

There is an ongoing exploration of different methods to approach the problem. Some participants have provided hints about using conservation of energy and the relationship between kinetic and potential energy. However, there is no clear consensus on the correct path forward, and multiple interpretations of the problem are being discussed.

Contextual Notes

Participants note the complexity of the problem and express frustration over previous attempts that did not yield correct results. There is a focus on deriving expressions for energy as functions of time, but some participants are struggling with the mathematical manipulations required.

  • #61
Recheck your arithmetic.
 
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  • #62
I have twice and i still get the same answer...where are you looking... or can you point it out because this questions has been driving me up the wall for almost two days. So if possible can you point out exactly where i went wrong because obviously you know that i know what i am doing.
 
  • #63
I suspect you are dropping an \omega.
 
  • #64
I don't believe i am...i am using the exact d(KE)/dt equation that i posted a few posts ago. So as far as i know i am not. What numeric value do you get.? If you choose to calculate it i will redo the question showing you exactly what i am doing.
 
  • #65
1.06

cos(3.70)(1.06)=-0.710627
sin(3.70)(1.06)=-0.703568948

coswtsinwt=0.499975091

so now that i manipulate the d(KE)/dt i get:

KA^2wcoswtsinwt
A=0.0455 m
K=3.76 N/m

(3.76)(2.07025e-3)(3.70)(-0.710627)(-0.703568948)
=0.0143999 J/s
 
  • #66
You are a good man DOC AI...a very good man

thank you so much for your help...i probably drove you up the wall with this question but you stood by me

truly a super mentor

:biggrin:
 

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