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Simple Harmonic Motion: Mass on a spring 
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#1
Nov107, 11:11 PM

P: 65

1. The problem statement, all variables and given/known data
A massless spring with spring constant 27.7 N/m hangs vertically. A body of mass 0.31 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released. How far below the initial position does the body descend? 2. Relevant equations F = kx 3. The attempt at a solution I realized that the weight is the force so I set kx = mg, however I was told that in actuality the equation should be 2mg = kx, so my question is why is it 2mg? I don't get it. 


#2
Nov107, 11:26 PM

Admin
P: 21,887

kx = mg would be the situation in a static case, i.e. not initial movement.
In problem given, the mass is allowed to fall with the spring unstretched, so the mass will fall past is equilibrium position, reach a point where it stops, then rebound. If not damping is present, the mass would oscillate according to SHM. Now in equilibrium, kx = mg, so that deflection would represent the equilibrium position. Starting from the unstretched position, the mass falls a distance x to the equilibrium position and then must continue another displacement x until it stops (velocity equals zero) at which point the spring pulls back. http://hyperphysics.phyastr.gsu.edu...ermot2.html#c3 


#3
Aug2008, 08:28 PM

P: 21

hi astronuc, i still don't understand why it's 2mg



#4
Aug2008, 08:30 PM

P: 21

Simple Harmonic Motion: Mass on a spring
do you mean that F = kx only applies before the spring moves?
because i've measured spring constants by plotting Force vs. elongation graphs where force = mass hanging from spring *9.8 is that correct 


#5
Aug2008, 09:20 PM

P: 1,133

Its probably because they're not asking for its descent to equilibrium. As Astronuc said, what will happen when attaching the weight is that the spring will first move towards some maximum displacement, then move backwards towards its equilibrium position. What they're probably asking for is the maximum displacement.



#6
Aug2008, 09:44 PM

P: 152

I think what Astro said is the easiest way to get to the answer. He finds the delta x to go from initial position to equilibrium. Then applies the symmetry of harmonic motion to get the final answer.
Alternately, here is one way to get the 2mg=kx that you are asking about. In the initial position (unstretched string, mass hanging on it), what is the sum of the forces? Use F=ma to find a and call it a1. Now the mass falls to the equilibrium (a=0) point and continues down to the bottom of the motion. At that point, what is the sum of the forces? Use F=m*a2. Again apply the symmetry of the problem. What is the relation between a1 and a2? Plug in a1 for a2 and what do you get? Basically what 2mg=kx (=Spring Force) is saying is that the spring is pulling the mass back up at a force equal to twice it's weight. 


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