Energy Conservation as an explanation for v=0 for spring mass system

[User1265]

Homework Statement

I watched a video on a wesbite which explained the graph attached below for the following question:


A block of mass m is connected to the ceiling by a spring with spring constant k. Initially the spring has zero extension and the block is held in position by a support. At time t=0 the support is removed from the block and it falls vertically downwards. Given that the spring has no damping, and ignoring air resistance, draw a graph of the vertical position of the block as a function of time during its initial fall and oscillation.

Relevant Equations

F= -kx

My Solution:

For the displacement graph, the gradient is crucial to predict the behaviour of the displacement of the block through time.

At 1: System is released - velocity is zero, considering forces acting on block, kx < mg, as block is observed to move downwards, and object is accelerating in the negative direction, taking direction of mg to be -ve

At 2: Velocity is increasing so graphs gradient can be seen to steepen in the negative direction.

(Velocity is increasing at a decreasing rate - by virtue of springs extension, restoring force is increasing linearly in the opposite direction, hence net acceleration in direction of motion is decreasing (to zero) - thus there must come a time when resotring force is equal in size to the objects weight.)At 3 : Net acceleration in -ve direction approaches zero, kx = mg , then the block at equilbrium moves with a constant velocity , as seen by green straight line, representing velocity not changing.

At 4: Due to object's inertia , continues past equilbrium, and as block contunues to move below equilibrum, its restoring force is still increasing so now velocity in that direction decreasing as acceleration is now increasing in the opposite (+ve) direction as kx > mg

At 5: in negative diretcion the decreasing velocity is approaching zero, (as kx>>mg) hence curve flattens out

Part Where I am stuck

At 6: The block reaches a point of lowest displacement by virtue of the fact the velocity equals zero, at this point in time, (the extension of spring has reached its maximum with the given weight and so the net acceleration upwards must be at a maximum).

This was my reasoning since velocity is decreasing to zero in negative direction there cannot be any further displacement thereon from this point in time. Hence the block can be said to reach its maximum displacement.

I thought by reasoning this way it would at least be consistent with my considerations of velocity and acceleration in turn affecting velocity in the previous parts.

However the actual solution At 6 gives the reason of conversation of energy as reasoning for why v=0 - stating block moves downwards until a point where all Kinetic energy and gravitational potential energy transferred to elastic PE in the spring, so at this point of maximum displacement the block must have zero velocity.

Q: I don't understand why the argument has shifted to the conservation of energy at the critical point where velocity reaches zero at the maxmium displacement, as opposed to just carrying on the consideratons of velocity decreasing until it is about to change direction. Is it better, more intuitive explanation?

 

[PeroK]

Homework Statement: I watched a video on a wesbite which explained the graph attached below for the following question:A block of mass m is connected to the ceiling by a spring with spring constant k. Initially the spring has zero extension and the block is held in position by a support. At time t=0 the support is removed from the block and it falls vertically downwards. Given that the spring has no damping, and ignoring air resistance, draw a graph of the vertical position of the block as a function of time during its initial fall and oscillation.
Homework Equations: F= -kx

View attachment 253224
My Solution:

For the displacement graph, the gradient is crucial to predict the behaviour of the displacement of the block through time.

At 1: System is released - velocity is zero, considering forces acting on block, kx < mg, as block is observed to move downwards, and object is accelerating in the negative direction, taking direction of mg to be -ve

At 2: Velocity is increasing so graphs gradient can be seen to steepen in the negative direction.

(Velocity is increasing at a decreasing rate - by virtue of springs extension, restoring force is increasing linearly in the opposite direction, hence net acceleration in direction of motion is decreasing (to zero) - thus there must come a time when resotring force is equal in size to the objects weight.)At 3 : Net acceleration in -ve direction approaches zero, kx = mg , then the block at equilbrium moves with a constant velocity , as seen by green straight line, representing velocity not changing.

At 4: Due to object's inertia , continues past equilbrium, and as block contunues to move below equilibrum, its restoring force is still increasing so now velocity in that direction decreasing as acceleration is now increasing in the opposite (+ve) direction as kx > mg

At 5: in negative diretcion the decreasing velocity is approaching zero, (as kx>>mg) hence curve flattens out

Part Where I am stuck

At 6: The block reaches a point of lowest displacement by virtue of the fact the velocity equals zero, at this point in time, (the extension of spring has reached its maximum with the given weight and so the net acceleration upwards must be at a maximum).

This was my reasoning since velocity is decreasing to zero in negative direction there cannot be any further displacement thereon from this point in time. Hence the block can be said to reach its maximum displacement.

I thought by reasoning this way it would at least be consistent with my considerations of velocity and acceleration in turn affecting velocity in the previous parts.

However the actual solution At 6 gives the reason of conversation of energy as reasoning for why v=0 - stating block moves downwards until a point where all Kinetic energy and gravitational potential energy transferred to elastic PE in the spring, so at this point of maximum displacement the block must have zero velocity.

Q: I don't understand why the argument has shifted to the conservation of energy at the critical point where velocity reaches zero at the maxmium displacement, as opposed to just carrying on the consideratons of velocity decreasing until it is about to change direction. Is it better, more intuitive explanation?

Your explanation was good up to the point where you said you got stuck. Why not continue with this line of reasoning?

In general, you can look at a problem like this through analysis of forces and/or energy conservation. In this case, looking at forces seems to me to be the best way to explain the different phases of the motion.

 

[haruspex]

It depends what you are trying to explain. If asking why it reaches a lowest point you cannot answer that it is because the velocity becomes zero. You would then have to prove that it does so.
The conservation of energy argument doesn’t really work either. It supposes an initial finite total of energy then reasons that if it were to descend far enough then the elastic energy would hold it all, leaving nothing for KE. To see that this flawed, imagine gravity increasing as the mass descends, maybe towards a black hole.