What Is the Maximum Velocity of an Object in Harmonic Motion?

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SUMMARY

An object with a mass of 3.5 kg attached to a spring with a stiffness constant of 300 N/m is undergoing simple harmonic motion. The object, when 0.020 m from its equilibrium position, has a speed of 0.55 m/s. To calculate the maximum velocity, one must apply the conservation of energy principle, combining kinetic and potential energy equations. The maximum velocity occurs when potential energy is zero, allowing the total energy to convert entirely into kinetic energy.

PREREQUISITES
  • Understanding of simple harmonic motion
  • Knowledge of conservation of energy principles
  • Familiarity with kinetic and potential energy equations
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of energy conservation in harmonic motion
  • Learn how to calculate potential energy in spring systems
  • Explore maximum velocity calculations in oscillatory systems
  • Investigate the effects of mass and spring constant on oscillation characteristics
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts of energy conservation in harmonic systems.

chazgurl4life
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An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 300 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.

b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]

so if i use E= 1/2mv^2=1/2KX^2
then-->.5(3.5 kg)(.55) ^2= 1.5(300 N/m)(.200) ^2

but what does that prove ? I don't think this is right
 
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chazgurl4life said:
An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 300 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.

b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]

so if i use E= 1/2mv^2=1/2KX^2
then-->.5(3.5 kg)(.55) ^2= 1.5(300 N/m)(.200) ^2

but what does that prove ? I don't think this is right
You are right - your answer is not right.

Just add the kinetic and potential energy to find its total energy. Energy is conserved. So when potential energy = 0, kinetic energy = total energy. That is where it will have maximum speed.

AM
 
chazgurl4life said:
When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.
That piece of information let's you calculate the potential and kinetic energies (at the point .02m from equillibrium). With that you can calculate the total energy of the system(E = P+K). Now how is this connected to the maximum velocity the object can have?
 

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