Find the period of the motion and the farthest distance from equilibrium

Click For Summary
SUMMARY

The discussion focuses on determining the period of motion and the maximum distance from equilibrium for an object subjected to forces described by Fx = -kx and Fy = -ky. The period is confirmed to be T = 2π√(m/k), indicating simple harmonic motion (SHM). The potential energy is expressed as U = 0.5kr², with the maximum distance from equilibrium dependent on the initial conditions. The conversation also addresses the scenario where the spring constants kx and ky differ, leading to two distinct periods for motion in the x and y directions.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with potential energy in conservative forces
  • Knowledge of mass-spring systems and their equations of motion
  • Basic calculus for integrating force functions
NEXT STEPS
  • Explore the implications of differing spring constants in multi-dimensional SHM
  • Study the derivation and application of the potential energy function U = 0.5kr²
  • Learn about energy conservation principles in oscillatory motion
  • Investigate the effects of initial conditions on oscillation characteristics
USEFUL FOR

Students of physics, particularly those studying mechanics and oscillatory systems, as well as educators seeking to clarify concepts of simple harmonic motion and energy conservation.

discordplus
Messages
9
Reaction score
0

Homework Statement



an object of mass m experiences a force of the form
Fx = -kx
Fy = -ky

Its initial speed is v and at that time it is at (x,y) = (0,0)

Find the period of the motion and the farthest distance from equilibrium

Homework Equations



U = -[tex]\int F(r) dr[/tex]
T = 2[tex]\pi[/tex][tex]\sqrt{m/k}[/tex]

The Attempt at a Solution


I think this is a superposition of SHM, so the period would just be 2pi root(m/k), yes?

Furthermore since the force acting on the particle is conservative it has a potential energy given by .5kr^2 and its initial energy is .5mv^2, so when its speed equals zero the potential energy is maxed so .5kr^2 = .5mv^2? v is again the initial speed

What happens if the k's in the force equation aren't equal though? In terms of the period i mean. Is the motion just characterized by two separate periods? One in x and one in y?
 
Physics news on Phys.org
discordplus said:
I think this is a superposition of SHM, so the period would just be 2pi root(m/k), yes?
'Sounds reasonable to me. :approve:
Furthermore since the force acting on the particle is conservative it has a potential energy given by .5kr^2 and its initial energy is .5mv^2, so when its speed equals zero the potential energy is maxed so .5kr^2 = .5mv^2? v is again the initial speed
That also sounds reasonable. :approve:

But using the above relationship, the exact maximum distance from equilibrium is contingent on the fact that at some point in the oscillation, it passes through point (0,0). That is specified in the problem statement. If that tidbit of information wasn't given, there wouldn't be enough information to solve the problem (without giving at least some sort of initial condition -- and even then, it might not be solved in exactly the same way). For example, the mass might just be going around in a circle (with a constant speed and a constant distance).
What happens if the k's in the force equation aren't equal though? In terms of the period i mean. Is the motion just characterized by two separate periods? One in x and one in y?
Yes, that's right. And you have to be a little more careful too about how you define your U and T equations, because there are now two ks to be concerned with, instead of just one. So you won't be able to factor out the k.
 

Similar threads

Replies
25
Views
1K
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
3
Views
991
Replies
4
Views
1K
Replies
24
Views
4K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 26 ·
Replies
26
Views
6K