Solving Simple Harmonics: Return Time After Collision

  • Thread starter Thread starter kevinr
  • Start date Start date
  • Tags Tags
    Harmonics
AI Thread Summary
A 10.0 kg mass collides with a stationary 10.0 kg mass attached to a spring, and the problem involves determining the time it takes for the system to return to its initial position after the collision. The key point is that all oscillations will be identical, defined by the energy imparted by the moving mass. The period of oscillation (T) can be used to find the time for the first return to the original position, which is less than T. The discussion emphasizes understanding the relationship between frequency and period while noting that the first return occurs before the completion of a full oscillation. The problem can be approached in multiple ways, with both straightforward and complex methods available for solving it.
kevinr
Messages
32
Reaction score
0
[SOLVED] Simple Harmonics

Homework Statement



A 10.0 kg mass is traveling to the right with a speed of 2.40 m/s on a smooth horizontal surface when it collides with and sticks to a second 10.0 kg mass that is initially at rest but is attached to a light spring with force constant 75.0 N/m.

How long does it take the system to return the first time to the position it had immediately after the collision?

Homework Equations



?

The Attempt at a Solution



I am sort of confused about this. I found the f, A, T for subsequent oscillations but what's the difference between them and the first oscillation as asked by this question?

Thanks!
 
Physics news on Phys.org
Hello,

No difference. Say the mass is coming in from the right, it comes in and collides with the mass on the spring (which I'm assuming here is in equilibrium, position defined as x = 0). It sticks, and the moving mass has imparted some energy, which is going to displace the mass on the spring, which it has now stuck to. So the whole thing is going to compress the spring some amount, come to a stop, and then the spring is going to push it back out again, at which time it will again reach (and overshoot, of course) x = 0.

The problem wants to know what time that will occur. All oscillations will be the same, and will be defined by the (kinetic) energy the moving mass imparted onto the stationary mass/spring system.

Hope that helps. If you know f or T, you can find the answer, but you have to be careful about how they're defined. There are a few (fun!) ways to solve this problem.
 
"There are a few (fun!) ways to solve this problem." :D

I can't any find of the fun ways to do this problem. I know f and T but what do i do with that?

(im thinking that it since my T = 3.24, it should be less) right?
 
Last edited:
Hello,

Well, frequency and period are inverses(T = \frac{1}{\nu}), so pick your poison-- I'd use period, since its already in seconds.

What is the period? Well, its the time that the system takes to complete one whole oscillation-- to compress the spring, spring back out, and come back to x=0. The tricky part, of course, is the fact that, assuming the mass was at x=0 at t=0, the mass will be at x=0 at t = T-- but this will not be the first time it is. Why not? You have to think about what's happening.

The hard way to solve the problem would be to find x(t) and calculate the first non-trivial x=0. Harder way, at least, IMHO. The first way I think is more clever, at least.

But both are fun!

Edit: PS: You're right, it will be less.
 
Ah ok thank you i got it now!
 
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...

Similar threads

Back
Top