Back to Contents

We don't have an obvious solvent for this integral. We could multiply and divide the integrand by x and try to integrate the result by parts, but I want to try something different. If we could only multiply the integrand by x, we would have

(Eq'n 1)

But we also know that

(Eq'n 2)

so can we combine those equations in a way that gives us a solution?

In their current form those equations don't give us a solution, but if we interchange the a2 and the x2 under the radicals, which has the effect of multiplying the radicals by i, we get

(Eq'n 3)

and

(Eq'n 4)

Multiplying Equation 4 by i and adding the result to Equation 3 gives us

(Eq'n 5)

which leads us directly to

(Eq'n 6)

If we now re-transpose a2 and x2 under the radicals, which has the effect of multiplying the radicals by -i (because we already multiplied them by +i when we first did the transposition on them), and then multiply the entire equation by -i, we get

(Eq'n 7)

Evaluating the right side of that equation from x=0 to x=X gives us

(Eq'n 8)

To evaluate the negative i-th power of the argument of the logarithm we recall that we can represent any complex number z=r+is, by way of de Moivre's theorem and Euler's formula, as

(Eq'n 9)

With that formula we can calculate

(Eq'n 10)

But in the argument of the logarithm in Equation 8 we have, by way of Equation 9, ρ=0 and Sinσ=X/a, so we can rewrite Equation 8 as

(Eq'n 11)

subject to the proviso that a>X.

Example Application:

Oscillation Period of a Mass on a Spring

We have a small body of mass m attached to one end of a spring of force constant k (F=-kx), which spring's other end attaches to an immovable frame. At static equilibrium the body occupies the point x=0, so the body's potential energy vis-a-vis the spring comes out as

(Eq'n 12)

If we then pull the body to the point x=x0, giving it a total energy of

(Eq'n 13)

and release it, it will oscillate about the point x=0 with some period P.

In order to determine that period we first calculate the body's speed at any point x from the body's kinetic energy (T=E-U) and we get

(Eq'n 14)

But we know that v=dx/dt, so we have

(Eq'n 15)

If we integrate the right side of that equation from x=x0 to x=-x0, then we also integrate the left side from t=0 to t=P/2, so we have

(Eq'n 16)

We then have the period

(Eq'n 17)

So now we can describe the motion of the body as

(Eq'n 18)

in which the angular frequency comes to us as

(Eq'n 19)

efefefaaabbbefefef

Back to Contents