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- ηtx(t)=0 .
Where m is the mass, b the damping constant, k and η positive constants and x(t) displacement of the spring from equilibrium position. Let m=1 kg, b=2 Nsec/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+1/2t3-1/4 t4+...
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)= Σ n=0∞ an xn
As given,
mx"(t)+bx'(t)+ke-ηt x(t)=0
Let
x(t)= Σ n=0∞ an tn
x'(t)= Σ n=1∞ nan tn-1
x''(t)= Σ n=2∞ n(n-1) an tn-2
The Maclaurin series is,
e-t= Σ n=0∞ (-t)n/n!
= Σ n=0∞ (-1)n (t)n/n!
Replace this in the equation.
Σ n=2∞ n(n-1) an tn-2+2 Σ n=1∞ nan tn-1+Σ n=0∞ (-1)n (t)n/n! =0
You will set coefficients equal to zero. The expression is,
2a2+2a1+a0 =0
a2= - (2a1+a0)/2
Hence, the expression is,
2a2+2a1+a0 =0
a2= - (2a1+a0)/2
Now you will find the coefficient.
a2= - (2a1+a0)/2
=-(2 (0)+1)/2
=-1/2
Now,
6a3+4a2-a0+a1=0
a3= (-4a2+a0-a1)/6
a3 =[-4 (-1/2)+1-0]/6
a3 =1/2
And
+12a4+6a3+(1/2)a0-a1+a2=0
a4=-[-6a3-(1/2)a0+a1+a2]/12
=[-6(1/2)-(1/2)(1)+0-(-1/2)]/12
=-1/4
Substitute the coefficients in the expression.
x(t)=1-1/2 t2+ 1/2 t3-1/4 t4+...
Hence, these are the first four nonzero terms.