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Investigation The differential equation 8 32 y " + b y ' + k y = 8 32 F ( t ) , y ( 0 ) = 1 2 , y ' ( 0 ) = 0 models the oscillating motion of an object on the end of aspring, where y is the displacement from equilibrium (positivedirection is downward), measured in feet, t is time in seconds, b is the magnitude of the resistance to the motion, k is the spring constant from Hooke’s Law, and F ( t ) is the accelerationimposed on the system. (a) Solve the differential equation and use a graphing utility tograph the solution for each of the assigned quantities for b , k , and F ( t ). b = 0 , k = 1 , F ( t ) = 24 sin π t b = 0 , k = 2 , F ( t ) = 24 sin ( 2 2 t ) b = 0.1 , k = 2 , F ( t ) = 0 b = 1 , k = 2 , F ( t ) = 0 (b) Describe the effect of increasing the resistance to motion b . (c) Explain how the motion of the object changes when astiffer spring (greater value of k ) is used.

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Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
Chapter 16, Problem 56RE
Textbook Problem
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Investigation The differential equation

8 32 y " + b y ' + k y = 8 32 F ( t ) , y ( 0 ) = 1 2 , y ' ( 0 ) = 0

models the oscillating motion of an object on the end of aspring, where y is the displacement from equilibrium (positivedirection is downward), measured in feet, t is time in seconds, b is the magnitude of the resistance to the motion, k is the spring constant from Hooke’s Law, and F(t) is the accelerationimposed on the system.

(a) Solve the differential equation and use a graphing utility tograph the solution for each of the assigned quantities for b, k, and F(t).

b = 0 , k = 1 , F ( t ) = 24 sin π t

b = 0 , k = 2 , F ( t ) = 24 sin ( 2 2 t )

b = 0.1 , k = 2 , F ( t ) = 0

b = 1 , k = 2 , F ( t ) = 0

(b) Describe the effect of increasing the resistance to motion b.

(c) Explain how the motion of the object changes when astiffer spring (greater value of k) is used.

a)

To determine

To calculate: The solution of the differential equation, 832y''(t)+by'(t)+ky(t)=832F(t). Also graph the solution for each of the assigned quantities for b, k, F(t) using graphing utility.

b=0,k=1,F(t)=24sinπt

b=0,k=2,F(t)=24sin(22t)

b=0.1,k=2,F(t)=0

b=1,k=2,F(t)=0

Explanation of Solution

Given information:

The given differential equation is 832y''(t)+by'(t)+ky(t)=832F(t), y(0)=12and y'(0)=0.

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:

(i)

Consider the given differential equation is 832y''(t)+by'(t)+ky(t)=832F(t).

The coefficients of the differential terms are,

b=0k=1F(t)=24sinπt

Now putting these in the differential equation

832y''(t)+(0)y'(t)+1y(t)=832(24sinπt)832y''(t)+y(t)=6sinπt

The characteristics equation is 832m2+1=0.

First, we have to find the solution of the characteristics equation 14m2+1=0.

14m2+1=014m2=1m2=4m=±4m=±2i

Therefore,

yh=C1cos2t+C2sin2t

Let yp be generalized form of 6sinπt .

yp=Acosπt+Bsinπtyp'=πAsinπt+πBcosπtyp''=π2Acosπtπ2Bsinπt

Substituting in the original equation yields

14y''(t)+y(t)=6sinπt14(π2Acosπtπ2Bsinπt)+Acosπt+Bsinπt=6sinπtπ24Acosπtπ24Bsinπt+Acosπt+Bsinπt=6sinπt3π24Acosπt+3π24Bsinπt=6sinπt3π24Acosπt+3π24Bsinπt=6sinπt

Equating the coefficient of like terms yields

B=8π2A=0

Therefore, yp=8π2sinπt and the general solution is ,

y=yh+ypy=C1cos2t+C2sin2t+8π2sinπt

Substituting the given initial conditions,

At, t=0,y(0)=12, therefore,

y=C1cos2t+C2sin2t+8π2sinπt12=C1cos(2×0)+C2sin(2×0)+8π2sin(π×0)12=C1

Now differentiating the general solution,

y=C1cos2t+C2sin2t+8π2sinπty'=2C1sin2t+2C2cos2t+8πcosπt

Substituting the given initial conditions,

At, t=0,y'(0)=0, therefore,

0=2C1sin(2×0)+2C2cos(2×0)+8πcos(π×0)0=2C2+8π2C2=8πC2=4π

Therefore,

C1=12

C2=4π

Therefore, the general solution of the given differential equation is y(t)=12cos2t4πsin2t+8π2sinπt.

Graph:

The graph of the differential equation y(t)=12cos2t4πsin2t+8π2sinπt is given below,

ii) Consider the given differential equation is 832y''(t)+by'(t)+ky(t)=832F(t).

The coefficients of the differential terms are,

b=0k=2F(t)=24sin(22)t

Now putting these in the differential equation

832y''(t)+(0)y'(t)+2y(t)=832(24sin(22t))832y''(t)+2y(t)=6sin(22t)

The characteristics equation is 832m2+2=0.

First, we have to find the solution of the characteristics equation 14m2+2=0.

14m2+2=014m2=2m2=8m=±22i

Therefore,

yh=C1cos(22t)+C2sin(22t)

Let yp be generalized form of 6sin(22t) .

yp=Acos(22t)+Bsin(22t)yp'=22Asin(22t)+22Bcos(22t)yp''=8Acos(22t)8Bsin(22t)

Substituting in the original equation yields

14y''(t)+2y(t)=6sin(22t)14(8Acos(22t)8Bsin(22t))+2Acos(22t)+2Bsin(22t)=6sin(22t)8Acos(22t)8Bsin(22t)+2Acos(22t)+2Bsin(22t)=6sin(22t)6Acos(22t)6Bsin(22t)=6sin(22t)Acos(22t)tBsin(22t)t=sin(22t)

Equating the coefficient of like terms yields

B=1A=0

Therefore, yp=sin(22t) and the general solution is ,

y=yh+ypy=C1cos(22t)+C2sin(22t)sin(22t)

Substituting the given initial conditions,

At, t=0,y(0)=12, therefore,

y=C1cos(22t)+C2sin(22t)sin(22t)12=C1cos(22×0)+C2sin(22×0)sin(22×0)12=C1

Now differentiating the general solution,

y=C1cos(22t)+C2sin(22t)sin(22t)y'=22C1sin(22t)+22C2cos(22t)22cos(22t)

Substituting the given initial conditions,

At, t=0,y'(0)=0, therefore,

y'=22C1sin(22t)+22C2cos(22t)22cos(22t)0=22C1sin(22×0)+22C2cos(22×0)22cos(22×0)0=22C22222C2=22C2=1

Therefore,

C1=12

C2=1

Therefore, the general solution of the given differential equation is y(t)=12cos(22t)+sin(22t)sin(22t)y(t)=12cos(22t)

Graph:

The graph of the differential equation y(t)=12cos(22t) is given below,

b)

To determine

The effect of increasing resistance to the motion b.

c)

To determine

How the motion of the object changes when a stiffer spring (greater value of k) is used.

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Chapter 16 Solutions

Multivariable Calculus
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