
#1
Oct904, 09:36 AM

P: 25

A 50.0g hardboiled egg moves on the end of a spring with force constant . It is released with an amplitude 0.300 m. A damping force acts on the egg. After it oscillates for 5.00 s, the amplitude of the motion has decreased to 0.100 m.Calculate the magnitude of the damping coefficient . Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures
pls who can help me? thanx 



#2
Oct904, 10:46 AM

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PF Gold
P: 12,016

How should Newton's 2.law of motion look like?




#3
Oct904, 10:54 AM

P: 25

i think it is:
kxbv=ma 



#4
Oct904, 11:32 AM

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PF Gold
P: 12,016

damping coefficient
That's correct!
Now, what type of solutions have you learnt that this differential equation has? 



#5
Oct904, 11:55 AM

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P: 2,280

See it as
[tex] kx  b \frac{dx}{dt} = m\frac{d^2 x}{dt^2} [/tex] 



#7
Oct904, 12:06 PM

P: 25

v= dx/dt and a= d^2/dt^2




#8
Oct904, 12:08 PM

P: 25

but what is the answer of d^2/dt^2 then?




#9
Oct904, 12:08 PM

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PF Gold
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mlee:
Any progress at what sort of solutions your equation has? 



#10
Oct904, 12:12 PM

P: 25

uh not really...;(




#11
Oct904, 12:14 PM

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PF Gold
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Now, I'd like you try a solution of the form:
[tex]x(t)=Ae^{rt}[/tex] (A and r constants) What condition must be placed on "r" in order for this to be a solution. Please post your work. 



#12
Oct904, 12:27 PM

P: 25

Asin(wt)+Bcos (wt)




#13
Oct904, 12:29 PM

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PF Gold
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Your oscillator is NOT undamped; try my approach, and post your work. 



#14
Oct904, 03:06 PM

P: 25

Aebt/2mCos(ω't + φ)




#15
Oct904, 03:08 PM

P: 25

Ae^(bt/2m)*cos(w't+φ)




#16
Oct904, 03:10 PM

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PF Gold
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You lack a minus sign in your exponential!
Now, knowing a) The initial displacement and b)That the initial velocity is zero How can you determine [tex]A,\phi[/tex] Besides, what is your value of "w"? 



#17
Oct904, 03:21 PM

P: 25

ω = sqrt(k/m)
ω' = √((k/m)  (bē/4mē)) 



#18
Oct904, 03:26 PM

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PF Gold
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Now, so how does your initial conditions determine [tex]A,\phi[/tex]?



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