To my knowledge I assume:(adsbygoogle = window.adsbygoogle || []).push({});

Newton's second law of motion :

F = ma = mx''

Hooke's Law

F = ks where s is the distance displaced by the mass.

When a mass is attached to the spring, the new spring force is:

F = k(s+x)

While the downward force is still:

mg

If the two forces are equal and opposite, then:

mg = k(s+x) after simplifying

kx=0

***********************

My question is: From kx=0, how do I get:

F=kx

I know how this leads to the final equation mx''-kx=0

But do we just assume that kx is a force and set it equal to F. I just don't understand the reasoning.

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# 2nd Order Linear - Modeling Spring Oscillation

Loading...

Similar Threads - Order Linear Modeling | Date |
---|---|

A Solve a non-linear ODE of third order | Feb 20, 2018 |

A Solving linear 2nd order IVP non-constant coefficient | Jan 10, 2018 |

I Classification of First Order Linear Partial Differential Eq | Jan 2, 2018 |

I Question about second order linear differential equations | Aug 21, 2017 |

B First Order Non-Linear ODE (what method to use?) | Feb 14, 2017 |

**Physics Forums - The Fusion of Science and Community**