
#1
Oct807, 08:20 PM

P: 88

1. The problem statement, all variables and given/known data
Determine the natural frequency of the system in the figure (attached). Assume the pulleys are frictionless and are of negligible mass. 2. Relevant equations k_eq=k_1+k_2 (springs connected in parallel) k_eq=(1/k_1)+(1/k_2) (springs connected in series) omega=sqrt(k/m) 3. The attempt at a solution I am stuck on figuring out if the springs are connected in parallel, or in series. I think it is in parallel because when the mass is placed there, both springs get displaced by the same amount. Once I figure out if it's parallel of series I can then easily find k_eq and treat the two springs as one. The mass will undergo undamped simple harmonic motion with a frequency of sqrt(k_eq/m). I am afraid that I am on the wrong track...to think that I can simply combine the springs into one? Thanks for the help, I really appreciate it. 



#2
Oct807, 08:53 PM

P: 88

anyway, I wrote the force equation as (treating down as positive)
mg(k_1)x(k_2)x=0 mg(k_1+k_2)x=0 mg(k_eq)x=0 What should my next step be...? 



#3
Oct1207, 05:32 PM

P: 88

I don't think I can simply treat the two springs combined together in either series or parallel fashion...?




#4
Oct1207, 07:00 PM

P: 863

springs...connected in series or in parallel?
If you pull on the mass, the two pulleys would feel an equal force, right?
So at least you know their frequencies should be the same. And I think that you can't have a series combination of two springs that aren't directly connected. I have to go in like a minute, so I can't really think this through, but so far I haven't been able to come up with a set up where they aren't in a line and still count as being in series. 



#5
Oct1207, 07:23 PM

P: 88

Hmmm...I asked my prof, he suggested to draw a FBD on it...I guess that would be the mass. I'm gonna work on the basis that the length of the string (or wire) doesn't change, and it is massless also (so are the pulleys).




#6
Oct1207, 07:25 PM

P: 88

But hmm...when the mass goes up, both of the springs should be compressed, and when the mass goes down, both springs should be stretched...is it a series connection here?




#7
Oct1207, 07:45 PM

P: 88

I think maybe the system is equivalent to the one shown in the attachement here...? I was thinking since the string or rod is massless and rigid...it could be taken out entirely...? The same goes for the pulleys which are massless also...




#8
Oct1207, 08:03 PM

P: 863

I can't view the attachment for some reason.
But anyway, series = easier spring compression, parallel = harder So figure out whether it's easier or harder with two springs. I say it makes things harder because both pulleys are exerting the same force on two different springs. If it was one force on two springs, then it would make things easier. But this is confusing... :( 



#9
Oct1207, 09:37 PM

P: 88

hmmm...ok i think the springs are connected in parallel...not too sure...but I am edging to parallel over series...




#10
Oct1207, 09:43 PM

P: 88

going with the parallel connection idea, k_eq=k_1+k_2
so the natural frequency of oscillation would be omega=sqrt((k_1+k_2)/m))... 



#11
Sep308, 08:57 AM

P: 1

I think it has to be parallel:
If you considered an equivalent system (taking mass/ friction of pulleys etc.) where you had a fixed pulley on the ceiling and moved the sprung pulley to the floor (I guess you'd need a second fixed pulley to have the mass still hanging downwards); the problem would seem a bit more clear. Also it looks like it would be twice as difficult to move the weight as it would with a singly sprung system (doubling the spring constant, rather than applying the series equation) Hope you agree 



#12
Mar611, 07:28 AM

P: 2

u can see the solution here..the value of keq suggests that u cannot directly predict whether the springs are in parallel or series b/c the pulley system is linked in terms of distance covered.



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