SHM: Vertical Displacement Above Equilibrium

[binbagsss]

SHM - VERTICAL

Surely whether you consider the motion used to establish the SHM system, or the motion as soon as the system is released you should derive the same equation.

DISPLACEMENT BELOW EQUILLBRIUM.
However, consider a verticle spring with particle attached hanging in equilibrium initially, displaced downward and subject to resistance.

-Considering intial conditions, acceleration is down => resistance is up, T is up (as spring extended), and mg down.
- However considering the moment the system is released, acceleration is up => resistance is down, T is up, mg is down.

The correct solution is the equation derived following the initial conditions.

(I note that for the conditions after, I have taken T as up, but this must only hold until length l is again reached, where compression may take over depending upon the relative elastic and gpe potential models).
(However I still do not understand why exactly conditions 2 would not mean in line to establish a valid shm equation, as some motion upward should still occur).

DISPLACEMENT ABOVE EQUILIBIRUM:

A system is initially in equilbrium it is the displaced in the direction upward from equilibrium - the displacement is such that the length of the spring is less than l. (for this example, l + e = 60 + 11.8cm) and displacement is 15cm.

- This time, considering conditions after release, leads me to dervie the correct equation. The particle will accelerate downward, mg acts downward and T acts upward.
- However, considering intial conditions, acceleration is upward, T would be downward (spring is now compressed) and mg would be downward.

Any assitance, immensely appreciated, thank you. =]

 

[anathema]

I can understand your confusion and the need for clarification on the derivation of the SHM equation in different scenarios. Let's break down the two scenarios you have mentioned - displacement below equilibrium and displacement above equilibrium.

In the first scenario, when the system is initially displaced below equilibrium and is subject to resistance, the forces acting on the particle are as you have correctly stated - acceleration is down, resistance is up, T is up (as spring is extended), and mg is down. This is in line with Newton's second law of motion, which states that the net force acting on an object is equal to its mass times its acceleration.

Now, when the system is released, the initial acceleration is up, which means that the net force acting on the particle is now in the upward direction. This results in the particle moving upward until it reaches its equilibrium position, where the forces acting on it are balanced and the net force is zero. At this point, the particle will continue to oscillate about its equilibrium position, with the spring providing the restoring force and the particle experiencing acceleration due to gravity.

In the second scenario, when the system is initially displaced above equilibrium, the forces acting on the particle are again in line with Newton's second law - acceleration is upward, T is downward (as the spring is compressed), and mg is downward. When the system is released, the initial acceleration is downward, resulting in the particle moving downward until it reaches its equilibrium position. Again, the forces acting on the particle will be balanced at this point and it will continue to oscillate about its equilibrium position.

In both scenarios, the correct SHM equation can be derived by considering the initial conditions and applying Newton's second law to the system. The equation will hold true for both the motion used to establish the SHM system and the motion after the system is released, as long as the initial conditions are taken into account.

I hope this helps clarify any confusion you may have had. Keep up the scientific curiosity and keep questioning the world around us!