   # Motion of a Spring In Exercises 31-34, use the differential equation w g y ' ' ( t ) + b y ' ( t ) + k y ( t ) = w g F ( t ) which models the oscillating motion of an object on the end of a spring (see figure). In the equation, y is the displacement from equilibrium (positive direction is downward), measured in feet, t is time in seconds, w is the weight of the object, g is the acceleration due to gravity, b is the magnitude of the resistance to the motion, k is the spring constant from Hooke’s Law, and F ( t ) is the acceleration imposed on the system. Find the displacement y as a function of time t for the oscillating motion described subject to the initial conditions. Use a graphing utility to graph the displacement function. w = 4 , g = 32 , b = 1 2 , k = 25 2 , F ( t ) = 0 y ( 0 ) = 1 2 , y ' ( 0 ) = − 4 ### Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378

#### Solutions

Chapter
Section ### Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
Chapter 16.3, Problem 34E
Textbook Problem
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## Motion of a Spring In Exercises 31-34, use the differential equation w g y ' ' ( t ) + b y ' ( t ) + k y ( t ) = w g F ( t ) which models the oscillating motion of an object on the end of a spring (see figure). In the equation, y is the displacement from equilibrium (positive direction is downward), measured in feet, t is time in seconds, w is the weight of the object, g is the acceleration due to gravity, b is the magnitude of the resistance to the motion, k is the spring constant from Hooke’s Law, and F(t) is the acceleration imposed on the system. Find the displacement y as a function of time t for the oscillating motion described subject to the initial conditions. Use a graphing utility to graph the displacement function. w = 4 , g = 32 , b = 1 2 , k = 25 2 , F ( t ) = 0 y ( 0 ) = 1 2 ,   y ' ( 0 ) = − 4

To determine

To calculate: The displacement y as a function of time t for the oscillating motion described subject to initial condition. Also graph the displacement function using graphing utility. The given differential equation is wgy''(t)+by'(t)+ky(t)=wgF(t) where w=4,g=32,b=12,k=252,F(t)=0.

### Explanation of Solution

Given information:

The given differential equation is wgy''(t)+by'(t)+ky(t)=wgF(t) where w=4,g=32,b=12,k=252,F(t)=0 and assume that y(0)=12and y'(0)=4.

Concept Used:

If yp is the particular solution of the differential equation y''+ay'+by=F(x) and yh is the general solution, then y=yh+yp is the general solution of the non homogeneous equation.

Calculation:

Consider the given differential equation is wgy''(t)+by'(t)+ky(t)=wgF(t).

The coefficients of the differential terms are,

wg=432=18b=12k=252F(t)=0

Now putting these in the differential equation

18y''(t)+12×y'(t)+252×y(t)=0

The characteristics equation is 18m2+12m+252=0m2+4m+100=0.

First, we have to find the solution of the characteristics equation m2+4m+100=0.

m2+4m+100=0m=4±42411002m=2±i96

Therefore,

yh=C1e2tcos96t+C2e2tsin96t

Let yp be generalized form of 0

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