   Chapter 17.3, Problem 3E

Chapter
Section
Textbook Problem

A spring with a mass of 2 kg has damping constant 14, and a force of 6 N is required to keep the spring stretched 0.5 m beyond its natural length. The spring is stretched 1 m beyond its natural length and then released with zero velocity. Find the position of the mass at any time t.

To determine

To find: The position of the mass at any time t .

Explanation

Given data:

The spring is stretched beyond its natural length, so x=0.5 , m=2kg , restoringforce=6N , damping constant=14

Formula used:

Write the expression for Hooke’s Law.

restoringforce=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 damping force.

dampingforce=cdxdt (2)

Here,

c is damping constant.

Write the expression for Newton’s Second Law.

md2xdt2+cdxdt+kx=0

mx+cx+kx=0 (3)

Write the expression for auxiliary equation.

mr2+cr+k=0 (4)

Write the expression for the  roots.

r1=cc24mk2m (5)

r2=c+c24mk2m (6)

Write the expression for general solution of over damping case.

x(t)=c1er1t+c2er2t (7)

Substitute 0.5m for x and 6N for restoring force in equation (1),

6N=k(0.5m)k=6N0.5mk=12Nm

Substitute 12 for k , 14 for c and 2 for m in equation (3),

2x+14x+12x=0

Find the auxiliary equation using equation (4).

2r2+14r+12=0

Find the value of r1 using equation (5),

r1=141424(2)(12)2(2)=14196964=141004=6

Find the value of r2 using equation (6),

r2=14+1424(2)(12)

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