Simple Harmonic Motion of a wave

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Homework Help Overview

The discussion revolves around a problem involving simple harmonic motion (SHM) as a cheerleader waves her pom-pom. The problem includes calculating maximum acceleration and velocity, as well as determining values at specific positions and the time taken to move certain distances.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various approaches to calculating maximum acceleration and velocity using given formulas. There is an exploration of how to approach parts (c) and (d) of the problem, with some suggesting the use of energy conservation principles.

Discussion Status

Some participants have provided insights into the energy approach for parts (a) and (b), while others express confusion about its application in part (d). Multiple interpretations of how energy relates to the problem are being explored.

Contextual Notes

Participants are navigating the constraints of the problem, particularly regarding the energy approach and its applicability to different parts of the question. There is a noted uncertainty about how time factors into the energy equations.

iwonde
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[SOLVED] Simple Harmonic Motion

Homework Statement


A cheerleader waves her pom-pom in simple harmonic motion with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is x = +9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach, and which ones cannot? Explain.


Homework Equations


abs(a_max)= ω^2A, A = amplitude
abs(v_max)=ωA
ω=2πf, f=frequency
a = -ω^2 sqrt(A^2 - x^2)



The Attempt at a Solution


(a) ω=2π(0.85)= 5.34 rad/s
abs(a_max)= (5.34)^2(0.18)= 5.13m/s^2
abs(v_max)= (5.34)(0.18) = 0.961 m/s

(b) a = -(5.34)^2 sqrt((0.18)^2 - (0.09)^2) = -4.45m/s^2
speed = abs(v) = (5.34) sqrt((0.18)^2 - (0.09)^2) = 0.832 m/s

I don't know how to approach (c) and (d).
 
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Maybe this will do for the (c) part:
x=As in\omega t.

(d) part seems easy. Just try calculating every part using the law of conservation:
E=E_{k}+E_{p}=E_{k(max)}=E_{p(max)}.
 
Last edited:
for d) I'm thinking that part (a) can be found by using the energy approach because K_max = (1/2)m(v_max)^2, from v_max, you can find a_max; part (b) cannot be found by using the energy method because energy doesn't deal with what happens in the beginning and end; part (c) cannot be found using the energy approach because the time variable doesn't occur in the energy equations.

Is that right?
 
The velocity can be found at any given point through conservation of energy.
 
I'm still a little confused about the energy approach in part (d).
 

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