Homework help: Harmonic Motion of a Spring and Block System

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SUMMARY

The discussion focuses on solving a harmonic motion problem involving a 0.1 kg block attached to a spring with a force constant of 17.2 N/m. The maximum speed of the block is calculated to be 2.09838 m/s using the formula Vm = Aω, where A is the amplitude of 0.16 m and ω is the angular frequency derived from ω = √(k/m). For subsequent parts, the user seeks to determine the block's speed at 8 cm from the equilibrium position and the time taken to move from 0 cm to 8 cm amplitude, utilizing the equations for velocity and acceleration in harmonic motion.

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  • Understanding of harmonic motion principles
  • Familiarity with the equations of motion for springs
  • Knowledge of angular frequency calculations
  • Ability to manipulate trigonometric functions in physics equations
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Students studying physics, particularly those focusing on mechanics and harmonic motion, as well as educators seeking to enhance their understanding of spring-mass systems.

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Homework Statement



Part 1:

A 0.1 kg block attached to a spring of force constant 17.2 N/m oscillates with an amplitude of 16 cm. Find the the maximum speed of the block.

I solved this part using the equations for angular frequency and max velocity.

Part 2:

Find speed of the block when it is 8 cm from the equilibrium position. Answer in m/s.

Part 3:

Find acceleration in the same situation ^

Part 4:

Find the time it takes the block to move from amplitude 0cm, to amplitude 8 cm.

Homework Equations



ω = [itex]\sqrt{}k/m[/itex]

Vm = Aω

V(t) = -AωSin(ωt)

a(t) = - Aω^2Cos(ωt)


The Attempt at a Solution



Part 1 i figured out:

ω = [itex]\sqrt{}17.2/0.1[/itex] = 13.1149 rads

Vm = (.16)(13.1149) = 2.09838 m/s


For part 2 i tried using the equation for velocity as a function of time. But I'm not sure how to solve the equation without the value for time. So far i have:

V(t) = -(0.08)(13.1149)Sin(13.1149(t))


Thank you everybody
 
Physics news on Phys.org
For part two, the displacement is half the amplitude so you are looking for a time [itex]t[/itex] such that [itex]\cos(\omega t) = 1/2[/itex].
 

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