Aging spring without damping. In a mass-spring system of aging spring discussed in Problem 30, assume that there is no damping (i.e., b=0), m=1 and k=1. To see the effect of aging consider as positive parameter.
(a) Redo Problem 30 with b=0 and η arbitrary but fixed.
(b) Set η =0 in the expansion obtained in part (a). Does this expansion agree with the expansion for the solution to the problem with η=0. [Hint: When η =0 the solution is x(t)=cos t].
(a) The first four nonzero terms are x(t)=1-1/2t2+η/6 t3+(1-η2)/24 t4+... .
(b) Yes, this expansion agree with the expansion with η =0.
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,
Taking derivative of the above equation,
The Maclaurin series is,
=∑n=0∞ (η)n/n! tη
Replace this in the equation.
∑n=2∞ n(n+1)antn-2+ ∑n=0∞ (η)n/n! tη ∑n=0∞ an tη=0
You will set coefficients equal to zero. The expression is,
Hence the expression is a2=-a0/2.
Now we have to find the coefficients.
(-η)2/2! a0+(-η)1/1! a1+(-η)0/0! a2=η2/2-ηa1+a2
a4= (-η2/2a0+η a1-a2)/12
Substitute the value of coefficients in the expression.
x(t)=1-1/2t2+η/6 t3+(1-η2)/24 t4+...
Hence the first four terms are x(t)=1-1/2t2+η/6 t3+(1-η2)/24 t4+... .
For η =0 the given equation is,
The solution is
From part (a) the solution will be,
Since this is a Taylor’s series for cos t the expansion in part (a) agrees with the expansion in part (b).
For Duffing's equation given in Problem 13, the behaviour of the solutions changes as r changes sign. When , the restoring force becomes stronger than for the linear spring. Such a spring is called hard. When , the restoring force becomes weaker than the linear spring and the spring is called soft. Pendulums act like soft springs.
(a) Redo Problem 13 with . Notice that for the initial conditions , the soft and hard springs appear to respond in the same way for small.
(b) Keeping and , change the initial conditions to and . Now redo Problem 13 with .
(c) Based on the results of part (b), is there a difference between the behavior of soft and hard springs for small? Describe.
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