   Chapter 17, Problem 20RE

Chapter
Section
Textbook Problem

A spring with a mass of 2 kg has damping constant 16, and a force of 12.8 N keeps the spring stretched 0.2 m beyond its natural length. Find the position of the mass at time t if it starts at the equilibrium position with a velocity of 2.4 m/s.

To determine

To find: The position of mass at time t if it is starting at equilibrium position with the velocity of 2.4m/s .

Explanation

Given data:

Mass (m) is 2 kg.

Damping constant (c) is 16.

Force (F) is 12.8 N.

Velocity is 2.4ms .

Formula used:

Consider the expression for the Hooke’s law.

F=kx (1)

Here,

F is force,

x is natural length, and

k is spring constant.

Substitute 12.8 N for F and 0.2 m for x in equation (1),

12.8N=k(0.2m)k=64

The damping constant k is known as positive constant. Then, the damping constant is 64.

Write the expression for the differential equation.

mx+cx+kx=0

Substitute 2 kg for m, 16 for c, and 64 for k,

2x+16x+64x=0 (2)

Consider the auxiliary equation.

2r2+16r+64=0

r2+8r+32=0 (3)

Roots of equation (3) are,

r=8±(8)24(1)(32)2(1){r=b±b24ac2afortheequationofar2+br+c=0}=8±i82=4±i4

Write the expression for the complementary solution of two complex roots r=α±iβ

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