   Chapter 17.3, Problem 4E

Chapter
Section
Textbook Problem

A force of 13 N is needed to keep a spring with a 2-kg mass stretched 0.25 m beyond its natural length. The damping constant of the spring is c = 8.(a) If the mass starts at the equilibrium position with a velocity of 0.5 m/s, find its position at time t.(b) Graph the position function of the mass.

(a)

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=0.25 , m=2kg , restoringforce=13N , damping constant=8

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.

r1r2}=c2m±ωi (5)

Here,

ω=4mkc22m (6)

Write the expression for general solution of under damping case.

x(t)=e(c2m)t(c1cosωt+c2sinωt) (7)

Substitute 0.25m for x and 13N for restoring force in equation (1),

13N=k(0.25m)k=13N0.25mk=52Nm

Substitute 52 for k , 8 for c and 2 for m in equation (3),

2x+8x+52x=0

Find the auxiliary equation using equation (4).

2r2+8r+52=0

Find the value of ω using equation (6).

Substitute 52 for k , 8 for c and 2 for m in equation (6),

ω=4(2)(52)(8)22(2)=3524=16×224=4224

Simplify ω as follows.

ω=22

Find the value of c2m .

c2m=82(2)=2

Substitute 22 for ω , and 2 for c2m in equation (7),

x(t)=e(2)t(c1cos22t+c2sin22t) (8)

Since, the spring starts at equilibrium the value of x(0)=0

(b)

To determine

To graph: The position function of the mass.

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