Damped harmonic oscillator for a mass hanging from a spring

  • #1
Phantoful
30
3

Homework Statement


g9XAO77.png


Homework Equations


Complex number solutions
z= z0eαt
Energy equations and Q (Quality Factor)

The Attempt at a Solution


For this question, I followed my book's "general solution" for dampened harmonic motions, where z= z0eαt, and then you can solve for α and eventually getting an answer of x=Ae-(ϒ/2)tcos(ω1t+∅) where ω1=sqrt((ω02-(ϒ/2)2)). This is just for the underdampened case and there are other solutions for the critical and overdampened case. However, I don't think these are the answers and I'm not even sure how to interpret these "general solutions". For this question would the case be any different if a v(0) = v0, and the mass is hanging? Should I treat it like a driven harmonic oscillator because of the hammer? This is the first time I'm answering a question like this one.
 

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  • #2
All the hammer blow does is to give you a non-zero initial velocity. Otherwise, just solve the ODE for the prescribed conditions.
 
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  • #3
Phantoful said:

Homework Statement


View attachment 223518

Homework Equations


Complex number solutions
z= z0eαt
Energy equations and Q (Quality Factor)

The Attempt at a Solution


For this question, I followed my book's "general solution" for dampened harmonic motions, where z= z0eαt, and then you can solve for α and eventually getting an answer of x=Ae-(ϒ/2)tcos(ω1t+∅) where ω1=sqrt((ω02-(ϒ/2)2)). This is just for the underdampened case and there are other solutions for the critical and overdampened case. However, I don't think these are the answers and I'm not even sure how to interpret these "general solutions". For this question would the case be any different if a v(0) = v0, and the mass is hanging? Should I treat it like a driven harmonic oscillator because of the hammer? This is the first time I'm answering a question like this one.

The general solution in the book is applicable to your problem. A hanging mass will oscillate about its equilibrium position. x is the deviation from the equilibrium.
Hitting once with the hammer does not mean that the oscillator is driven., It provides the initial conditions you have to fit the general solution to: At t=0 x(0)=0 and v(0)=v0. Determine A and θ for each case.
 
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  • #4
Thank you!
Dr.D said:
All the hammer blow does is to give you a non-zero initial velocity. Otherwise, just solve the ODE for the prescribed conditions.

ehild said:
The general solution in the book is applicable to your problem. A hanging mass will oscillate about its equilibrium position. x is the deviation from the equilibrium.
Hitting once with the hammer does not mean that the oscillator is driven., It provides the initial conditions you have to fit the general solution to: At t=0 x(0)=0 and v(0)=v0. Determine A and θ for each case.
 
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