Engery Conservation in Oscillatory Motion

[Julio Cesar]

A 0.980kg block slides on a frictionless, horizontal surface with a speed of 1.32m/s. The block encounters an unstretched spring with a force constant of 245N/m.

(a)How far is the spring compressed before the block comes to rest?

(b) How long is the block in contact with the spring before it comes to rest?



Equations: Conservation of energy using a spring.

E_i=E_f

1/2mv_o² = 1/2kA²

A= v_o√m/k

Question (b): T = 2∏√m/k




3. Attempt at solution shows:

A= 0.0835m

T= 0.397s

As far as the calculations, I know the amplitude as well as the period of one complete occilation. However, when the question asks me how long is the block in contact with the spring before it comes to rest...

Well, my senses tell me that it would be half a period since the spring is compressed when the block comes in contact with it and stops. (v= 0m/s)

The answer it gives me is 0.0993s...?!

Where does this come from; that's not half a period that's a quarter of a period. The answer may be staring at be straight in the face but I just don't get it. Can someone explain why the answer is 1/4 the period instead of 1/2 the period?

Thanks.

 

[ehild]

What is the time for zero displacement to maximum displacement during an SHM? http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html

ehild

 

[Julio Cesar]

Wow! I feel like an idiot now. I've just spent the last hour just looking at the diagram, wondering why and didn't even think to draw out a sine function, lol!

Thanks so much, this helps out a ton!

 

[ehild]

A drawing always helps.

ehild

 

[Nickie]

I can provide an explanation for why the block is in contact with the spring for only a quarter of a period instead of half a period. This has to do with the nature of oscillatory motion and the concept of "equilibrium".

When the block first comes in contact with the spring, it has a certain velocity (1.32m/s) and is moving towards the equilibrium point (where the spring is unstretched). As it compresses the spring, the spring exerts a force on the block in the opposite direction, causing the block to decelerate and eventually come to rest.

At this point, the block has reached the equilibrium point, where the spring is fully compressed and the block is not moving. This is the moment when the block has transferred all of its kinetic energy to the spring, and the spring has reached its maximum potential energy.

In order for the block to start moving in the opposite direction (towards the equilibrium point again), the spring needs to start releasing its potential energy and exerting a force on the block in the opposite direction. This is what causes the block to start moving again.

Now, if we consider the period of oscillation, it is the time it takes for the block to go from one equilibrium point to the other and back again. This means that the block is in contact with the spring for only half of the period, because the other half of the period is spent in motion away from the spring.

Therefore, the block is in contact with the spring for only a quarter of the period before it comes to rest, because it reaches its maximum potential energy at the equilibrium point and then starts moving away from the spring.

I hope this explanation helps to clarify why the block is in contact with the spring for only a quarter of the period instead of half. It is important to remember that in oscillatory motion, the concept of equilibrium plays a crucial role in understanding the behavior of the system.