Engery Conservation in Oscillatory Motion

AI Thread Summary
A 0.980kg block sliding at 1.32m/s encounters a spring with a force constant of 245N/m, leading to calculations for spring compression and contact time. The spring compresses to an amplitude of 0.0835m, and the period of oscillation is calculated as 0.397s. The block is in contact with the spring for 0.0993s, which is identified as a quarter of the period rather than half. This confusion arises from the understanding of displacement during simple harmonic motion (SHM), where the time to reach maximum displacement is indeed a quarter of the period. Visualizing the motion with a sine function clarifies the relationship between time and displacement in SHM.
Julio Cesar
Messages
5
Reaction score
0
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.

Ei=Ef

1/2mvo2 = 1/2kA2

A= vo√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.

 
Physics news on Phys.org
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!
 
A drawing always helps.

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top