Chapter 17.3, Problem 1E

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# A spring has natural length 0.75 m and a 5-kg mass. A force of 25 N is needed to keep the spring stretched to a length of 1 m. If the spring is stretched to a length of 1.1 m and then released with velocity 0, find the position of the mass after t seconds.

To determine

To find: The position of the mass after t seconds.

Explanation

Given data:

Initial velocity xâ€²(t)=0 , m=5â€‰kg , restoringâ€‰force=25â€‰N .

Formula used:

Write the expression for Hookeâ€™s Law.

restoringâ€‰force=âˆ’kx (1)

Here,

k is spring constant, and

x is difference between the natural length and length of due to force exerts.

Write the expression for Newtonâ€™s Second Law.

md2xdt2+kx=0

mxâ€³+kx=0 (2)

Write the expression for auxiliary equation.

mr2+k=0 (3)

Write the expression for the complex roots.

r=Â±iÏ‰ (4)

Here,

Ï‰=km

Write the expression for general solution of mr2+k=0 with complex roots (r) .

x(t)=c1cosÏ‰t+c2sinÏ‰t (5)

Find the value of x .

x=0.75â€‰mâˆ’1â€‰m=âˆ’0.25â€‰m

Substitute âˆ’0.25â€‰m for x and 25â€‰N for restoring force in equation (1),

25â€‰N=âˆ’k(âˆ’0.25â€‰m)k=25â€‰N0.25â€‰mk=100â€‰Nm

Substitute 100 for k and 5 for m in equation (2),

5xâ€³+100x=0

Find the auxiliary equation using equation (3).

5r2+100=0

Solve for r .

5r2=âˆ’100r2=âˆ’20r=âˆ’4Ã—5r=2âˆ’5

Simplify r as follows.

r=Â±25i (6)

Compare equation (4) and (6).

Ï‰=25

Substitute 25 for Ï‰ in equation (5),

x(t)=c1cos(25)t+c2sin(25)t (7)

Find the value of x(0)

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