Calculating Simple Harmonic Motion Parameters for a 10.0kg Mass

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SUMMARY

The discussion focuses on calculating parameters of Simple Harmonic Motion (SHM) for a 10.0 kg mass with an amplitude of 8.0 cm and a maximum acceleration of 3.26 m/s². Key equations include the relationship between angular frequency (ω), frequency (f), and period (T), specifically ω = 2πf and T = 1/f. The maximum speed occurs at the equilibrium position, while total mechanical energy is equivalent to kinetic energy at this position. The phase constant (φ) is deemed irrelevant for the calculations, as it only affects the initial position on the SHM curve.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM) principles
  • Familiarity with angular frequency (ω) and its relationship to frequency (f)
  • Knowledge of kinetic and potential energy in oscillatory systems
  • Ability to interpret and analyze SHM graphs
NEXT STEPS
  • Calculate the period (T) of the SHM using the maximum acceleration and amplitude
  • Determine the maximum speed of the mass at the equilibrium position
  • Compute the total mechanical energy of the system based on the mass and amplitude
  • Explore the effects of varying the phase constant (φ) on the SHM parameters
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to enhance their understanding of Simple Harmonic Motion calculations.

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Homework Statement


A 10.0kg mass undergoes Simple Harmonic Motion with an amplitude of 8.0 cm, a maximum acceleration of magnitude 3.26 m/s^2, and an unknown phase constant φ. What is the period, max speed of the mass, the total mechanical energy of the system, and the magnitude of the force on the particle when it is at half of its maximum displacement?


Homework Equations


a=xm-w^2cos(wt+φ)


The Attempt at a Solution


3.26=8.0-w^2cos(wt+φ)
what do i use for the w?
How can we do this without a phase.
 
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phase doesn't matters, its just where you set 0 time. So you would start at some arbitrary location on the SHM curve.
Maximum velocity occurs at the eqbm position.
total energy is equivalent to kinetic energy at eqbm
maximum acceleration, instantaneous, occurs when v=0, aka all energy is potential at the peak of the SHM

and

try drawing the SHM curve, it helps to understand what is going on
 
Last edited:
relevant equations
[itex]\omega = 2 \pi f[/itex]
[itex]T = \frac{1}{f}[/itex]
 

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