   # 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 = 2 , g = 32 , b = 1 , k = 4 , F ( t ) = 4 sin 8 t y ( 0 ) = 1 4 , y ' ( 0 ) = − 3 ### 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 33E
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 = 2 , g = 32 , b = 1 , k = 4 , F ( t ) = 4 sin 8 t y ( 0 ) = 1 4 ,   y ' ( 0 ) = − 3

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=2,g=32,b=1,k=4,F(t)=4sin8t.

### Explanation of Solution

Given information:

The given differential equation is wgy''(t)+by'(t)+ky(t)=wgF(t) where w=2,g=32,b=1,k=4,F(t)=4sin8t and assume that y(0)=14and y'(0)=3.

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=232=116b=1k=4F(t)=4sin8t

Now putting these in the differential equation

116y''(t)+1×y'(t)+4×y(t)=116×4sin8t116y''(t)+y'(t)+4×y(t)=14sin8ty''(t)+16y'(t)+64×y(t)=4sin8t

The characteristics equation is m2+16m+64=0.

First, we have to find the solution of the characteristics equation m2+16m+64=0.

m2+16m+64=0(m+8)2=0m=8

Therefore,

yh=(C1+C2t)e8t

Let yp be generalized form of 14sin8t

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