Find x(t) with Kx Force and Mass m | KxForce

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SUMMARY

The discussion focuses on solving the motion of a particle of mass m subjected to a force F(x) = kx, where K > 0. The initial conditions are given as the starting position x0 and an initial speed of zero. Participants suggest using the principles of integration and applying initial conditions to derive x(t) as a function of time. The conversation highlights the importance of correctly expressing acceleration and utilizing the relationship between velocity and position in the integration process.

PREREQUISITES
  • Understanding of Newton's second law: F = ma
  • Familiarity with basic calculus, specifically integration techniques
  • Knowledge of kinematics and motion equations
  • Concept of force as a function of position, specifically Hooke's Law: F(x) = kx
NEXT STEPS
  • Study the derivation of motion equations under constant forces
  • Learn about integrating functions with variable limits in calculus
  • Explore the application of initial conditions in differential equations
  • Investigate the relationship between velocity, acceleration, and displacement in physics
USEFUL FOR

Students studying classical mechanics, physics educators, and anyone interested in understanding the dynamics of particles under linear forces.

Ryan95
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Homework Statement


A particle of mass m is subject to force F(x)=kx with K>0. The initial starting position is x0 and the initial speed is zero. Find x(t).

Homework Equations


F(x)=kx
F=ma

The Attempt at a Solution


Honestly, I am totally lost on this. I've written acceleration as v(dv/dx) which gave me mv(dv/dx)=Kx and then tried separating variables to integrate, but once I do that, I'm totally lost as I end up with m(v2/2)=K(x2/2).
 
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Ryan95 said:
Honestly, I am totally lost on this. I've written acceleration as v(dv/dt) which gave me mv(dv/dx)=Kx and then tried separating variables to integrate, but once I do that, I'm totally lost as I end up with m(v2/2)=K(x2/2).

i wonder how one can write acceleration as v..dv/dt as we know it as rate of change of velocity with time.. may be a typo.
now dv/dt can be expressed as d/dxof v multiplied by dx/dt .

i think you should proceed with the analysis as per the rule of integration and have initial conditions at t=0 and try to find x as a function of t. as one normally does with constant forces.
 
drvrm said:
i wonder how one can write acceleration as v..dv/dt as we know it as rate of change of velocity with time.. may be a typo.
now dv/dt can be expressed as d/dxof v multiplied by dx/dt .

i think you should proceed with the analysis as per the rule of integration and have initial conditions at t=0 and try to find x as a function of t. as one normally does with constant forces.

Oh, thank you, yes that was a typo. I've edited the post.
 

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