   Chapter 17.3, Problem 2E

Chapter
Section
Textbook Problem

A spring with an 8-kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equilibrium position and is given an initial velocity of 1 m/s. 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=0.4 , Initial velocity x(t)=1ms ,

m=8kg , restoringforce=32N .

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 Newton’s Second Law.

md2xdt2+kx=0

mx+kx=0 (2)

Write the expression for auxiliary equation.

mr2+k=0 (3)

Write the expression for the complex roots.

r=±iω (4)

Here,

ω=km

Write the expression for general solution of mr2+k=0 .

x(t)=c1cosωt+c2sinωt (5)

Substitute 0.4m for x and 32N for restoring force in equation (1),

32N=k(0.4m)k=32N0.4mk=80Nm

Substitute 80 for k and 8 for m in equation (2),

8x+80x=0

Find the auxiliary equation using equation (3).

8r2+80=0

Solve for r .

8r2=80r2=10r=10

Simplify r as follows.

r=±10i (6)

Compare equation (4) with equation (6).

ω=10

Substitute 10 for ω in equation (5),

x(t)=c1cos(10)t+c2sin(10)t (7)

Since, the spring is start at its equilibrium position the value of x(0)=0

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