The Differential Equation of Free Motion

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SUMMARY

The discussion focuses on solving a differential equation for a spring-mass system with a spring constant (k) of 100 pounds per foot and a weight of 1 pound. The correct equation of motion is derived as (1/32)x'' + 100x = 0, with initial conditions x(0) = 1/12 feet and x'(0) = 0. The characteristic equation yields complex roots, leading to the general solution y(t) = A cos(10t) + B sin(10t). The period of the motion is determined using the relationship between angular frequency and period.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with spring-mass systems and Hooke's Law.
  • Knowledge of initial conditions and their role in solving differential equations.
  • Basic trigonometric functions and their applications in motion equations.
NEXT STEPS
  • Study the derivation of the characteristic equation for second-order differential equations.
  • Learn about the physical interpretation of spring constants and mass in oscillatory motion.
  • Explore the method of undetermined coefficients for solving non-homogeneous differential equations.
  • Investigate the relationship between angular frequency and period in harmonic motion.
USEFUL FOR

Students studying physics or engineering, particularly those focusing on mechanics and differential equations, as well as educators seeking to clarify concepts related to oscillatory motion and spring dynamics.

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Homework Statement


A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then stretched an additional 1 inch and released. Find the equation of motion, the amplitude, and the period. Neglect friction.

Homework Equations


mx"+kx=0

The Attempt at a Solution



I understand that I need to set up a differential equation but I'm not quite sure if this is right.

K = 100 lb/ft m = 1 lb

thus plugging into the equation

x" + 100x = 0

use characteristic equation,

I get r= +/- 10i

y^{g}(t)= cos (10t) + sin (10t)So far is everything is alright?

When I tried to solve for period. I set the inside, 10t=2pi. At least that's how my professor does it but apparently it's wrong when I submit my answers online.

I'm also not understanding how the initial conditions work?
 
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You need constants A and B in front of your trig terms to get the general solution. You are given the initial position is one inch (watch your units), plus or minus depending on your coordinates, and apparently the initial velocity is zero. Use these two facts to determine your A and B.
 
Thanks anyway but I solved this problem.

I rethought about the mass and realized that 1 pound is equal to mg. Thus 1/32 is the actual mass.

So the equation should have been

1/32x" + 100x=0, x(0)=1/12, x'(0)=0

So yeah, now everything makes sense.
 

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