Q30E
Question
Aging spring. As a spring ages, its “spring constant” decreases on value. One such model for a mass-spring system with an aging spring is
mx"(t)+bx'(t)+ke-ηt x(t)=0
where m is the mass, the damping constant, k and η positive constants and x(t) displacement of the spring from equilibrium position. Let m=1 kg, b=2 N sec/m, k=1 N/m, η =1 sec-1. The system is set in motion by displacing the mass 1m from it equilibrium position and releasing it (x(0)=1, x'(0)=0). Find at least the first four nonzero terms in a power series expansion of about t=0 of displacement.
Step-by-Step Solution
VerifiedThe first four nonzero terms in a power series expansion of about of displacement are x(t)=1-1/2t2+1/2t3-1/4t4+... .
The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a recurrence relation for the coefficients.
A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable.
It is generally given by the formula,
y(x)=Σn=0∞ anxn
Given,
mx"(t)+bx'(t)+ke-ηt x(t)=0
Let
x(t)=Σn=0∞ antn
Taking derivative of the above equation,
x'(t)=Σn=1∞ nantn-1
x"(t)=Σn=2∞ n(n-1)antn-2
The Maclaurin series is,
e-t = ∑n=0∞ (-t)n/n!
= ∑n=0∞ (-1)n tn/n!
Replace this in the equation.
∑n=2∞ n(n-1)antn-2+2 ∑n=1∞ nantn-1 +∑n=0∞ (-1)n tn/n! +∑n=0∞antn=0
You will set coefficients equal to zero. The expression is,
2a2+2a1+a0=0
a2= - (2a1+a0)/2
Hence the expression is a2= - (2a1+a0)/2.
Now you will find the coefficient.
a2= - (2a1+a0)/2
= -[2 (0)+1]/2
=-1/2
6a3+4a2-a0+a1=0
a3=(a0-a1 -4a2)/6
= -[4(-1/2)+1-0]/6
=1/2
12a4+6a3+1/2a0-a1+a2=0
a4= (-6a3-1/2a0+a1-a2)/12
=[-6(1/2)-1/2(1)+0-(-1/2)]/12
= -1/4
Substitute the coefficients in the expression.
x(t)=1-1/2t2+1/2t3-1/4t4+...
Hence, the first four nonzero terms are x(t)=1-1/2t2+1/2t3-1/4t4+....