Finding acceleration, velocity, and time for simple harmonic motion

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SUMMARY

The discussion focuses on calculating key parameters of simple harmonic motion (SHM) for a cheerleader's pom-pom with an amplitude of 0.180m and a frequency of 0.850Hz. The maximum velocity was determined to be 0.961m/s and the maximum acceleration was calculated as 5.13m/s² using the formulas for maximum velocity (vx,max = ωA) and maximum acceleration (ax,max = -ω²A). The time to move from the equilibrium position to a point 0.120m away was found to be 7.16s. The discussion also addresses the energy approach to find quantities related to velocity and acceleration.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM) principles
  • Familiarity with angular frequency (ω) calculations
  • Knowledge of derivatives in the context of motion equations
  • Proficiency in using energy conservation equations (E = K + U)
NEXT STEPS
  • Study the derivation and application of angular frequency in SHM
  • Learn how to apply derivatives to motion equations for velocity and acceleration
  • Explore energy conservation principles in mechanical systems
  • Investigate the relationship between amplitude, frequency, and maximum velocity in SHM
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and simple harmonic motion, as well as educators seeking to enhance their understanding of SHM calculations and energy approaches.

andreaumali
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Homework Statement



A cheerleader waves her pom-pom in simple harmonic motion with an amplitude of .180m and a frequency of .850Hz.

a) Find the maximum magnitude of the acceleration and of the velocity.

b) Find the acceleration and speed when the pom-pom's coordinate is x=+.090m.

c) Find the time required to move from the equilibrium position directly to a point .120m away.

d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach (E = K + U = 1/2 mv^2 + mgh = 1/2 kA^2) and which cannot? Explain.


Homework Equations



i. x=Asin(ωt)

ii. vx,max=ωA

iii. ax,max=-ω2A

where ω=angular frequency, A=amplitude, x=displacement from equilibrium, t=time

K=1/2 mv^2

U=mgh


The Attempt at a Solution



I was able to calculate (a) using formulas ii and iii. vmax=.961m/s; amax=5.13m/s^2

I was also able to calculate (c) using formula i. x=7.16s

However, I am unsure of how to find (b).

I think (d) is velocity but I'm not sure if from velocity you'd be able to get x and acceleration?
 
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For part b you know ω, A, and x. You should be able to solve for and find t.

I guess one way to solve "b" is by using your equation(i), you can find speed and acceleration by taking the 1st and 2nd derivative wrt t.
 

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