Simple Harmonic Motion: Mass on a spring

In summary, a massless spring with a spring constant of 27.7 N/m hangs vertically and a body of mass 0.31 kg is attached to its free end and released. The spring was unstretched before the body was released and the goal is to find how far below the initial position the body will descend. The equation F = kx is used, where F represents weight and kx represents the force of the spring. In equilibrium, kx = mg, but in this problem, 2mg = kx, because the mass will fall past the equilibrium position, stop, and then rebound. The final answer can be found by finding the displacement from the initial position to the equilibrium position, and then applying the symmetry
  • #1
delecticious
65
0

Homework Statement


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?


Homework Equations



F = kx

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.
 
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  • #2
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.phy-astr.gsu.edu/hbase/permot2.html#c3
 
  • #3
hi astronuc, i still don't understand why it's 2mg
 
  • #4
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
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
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.
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which a restoring force acts on an object towards a central equilibrium point. The motion follows a sinusoidal pattern and is characterized by a constant frequency and amplitude.

2. What is a mass on a spring?

A mass on a spring is a common physical system used to demonstrate simple harmonic motion. It consists of a mass attached to a spring, and when the spring is stretched or compressed, the mass will oscillate back and forth around the equilibrium point.

3. What is the restoring force in simple harmonic motion?

The restoring force in simple harmonic motion is the force that pulls an object back to its equilibrium position. In the case of a mass on a spring, the restoring force is provided by the spring itself, which exerts a force proportional to the displacement of the mass.

4. How does the mass affect the period of simple harmonic motion?

The mass does not affect the period of simple harmonic motion. The period, or the time it takes for one complete oscillation, is determined by the length of the spring and the force constant of the spring, not the mass of the object attached to it.

5. What factors can affect the amplitude of simple harmonic motion?

The amplitude of simple harmonic motion can be affected by the initial displacement of the mass, the force constant of the spring, and any external forces acting on the system. In general, a larger initial displacement or a weaker spring will result in a larger amplitude of motion.

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