Solving Simple Harmonics: Return Time After Collision

  • Thread starter Thread starter kevinr
  • Start date Start date
  • Tags Tags
    Harmonics
Click For Summary

Homework Help Overview

The problem involves a 10.0 kg mass colliding with another 10.0 kg mass attached to a spring, seeking to determine the time it takes for the system to return to its initial position after the collision. The subject area pertains to simple harmonic motion and collision dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between frequency and period, and how to apply these concepts to find the time for the system to return to the initial position. There is confusion regarding the distinction between the first oscillation and subsequent oscillations.

Discussion Status

Some participants have offered insights into the nature of oscillations and the definitions of period and frequency. There is an ongoing exploration of how to approach the problem, with multiple interpretations being considered.

Contextual Notes

Participants are navigating the definitions of oscillation parameters and the implications of the collision on the system's motion. There is an acknowledgment of the complexity involved in determining the first return time after the collision.

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([itex]T = \frac{1}{\nu}[/itex]), 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!
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
16
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
3
Views
988
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
1
Views
1K