   Chapter 17.3, Problem 1E

Chapter
Section
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 x(t)=0 , m=5kg , restoringforce=25N .

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 with complex roots (r) .

x(t)=c1cosωt+c2sinωt (5)

Find the value of x .

x=0.75m1m=0.25m

Substitute 0.25m for x and 25N for restoring force in equation (1),

25N=k(0.25m)k=25N0.25mk=100Nm

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

5x+100x=0

Find the auxiliary equation using equation (3).

5r2+100=0

Solve for r .

5r2=100r2=20r=4×5r=25

Simplify r as follows.

r=±25i (6)

Compare equation (4) and (6).

ω=25

Substitute 25 for ω in equation (5),

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

Find the value of x(0)

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