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Frequency of oscillation (spring) 
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#1
May1811, 08:59 PM

P: 42

1. The problem statement, all variables and given/known data
Two identical springs of spring constant 240 N/m are attached to each side of a block of mass 21 kg. The block is set oscillating on the frictionless floor. What is the frequency (in Hz) of oscillation? Figure: http://edugen.wiley.com/edugen/cours...5/fig15_30.gif 2. Relevant equations w = sq(k/m) w = omega = angular frequency k = spring constant m = mass w=2(pi)f f = w/(2pi) 3. The attempt at a solution So this problem seemed simple enough... w = sq(k/m) w = sq(240/21) w = 3.38 f = w/(2pi) f = 3.38/(2pi) = 0.538Hz Why is this wrong? It must have something to do with there being two springs... but I don't know how that would change things. 


#2
May1811, 09:12 PM

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P: 11,614

Suppose that if the block were at rest at the equilibrium point you were to displace it, say, towards the lefthand spring by a small amount x. Each spring is going to react with some force. What would be the net force required to make the displacement? How does that compare to what a single spring would do?



#3
May1811, 09:33 PM

P: 42

hmm... One would be pushing (KE?), and one would be pulling (PE?). Maybe I am to use energy conservation? There are no external forces, so then Emec is conserved. KE = 1/2* K (Xm)^{2} sin^{2}(wt + :theta:) U(t)= 1/2* k (Xm)^{2}cos^{2}(wt + :theta:) but it would be difficult to get "w" out of those functions. Is there a better way to do this? 


#4
May1811, 10:04 PM

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P: 11,614

Frequency of oscillation (spring)



#5
May1811, 10:30 PM

P: 42

If the forces are moving together then > 2F = kd so K will be half the size. So then I apply that to this equation: w = sq(.5k/21) = 2.39 f = w/(2pi) = (2.39/(2pi)) = 0.380 s^{1}?? 


#6
May1811, 10:44 PM

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P: 11,614

No, the force is doubled. F = 2Kd.
What, then, is the effective spring constant? 


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