## Wednesday, August 14, 2013

### Applications of Higher Order Differential Equations (2) Diagram for Problem 2 of Applications

We continue with further applications on the use of higher order differential equations to solve problems.

Example (1) Stretched spring: A 1 kg mass which naturally stretches a spring 5 cm is pulled down 5 cm farther and released. Find the equation of motion, its period and frequency.

Solution:  Note that: 5 cm = 0.05 m,

To get k, the spring constant, note: F = ma = -kx

A 1 kg mass exerts a downward force F = 10 N on the spring  (assuming g = 10 ms-2). Since:

F = ma = mg = (1 kg) (10 ms-2) = 10 kgms-2  =  10 N

So: k = 10N/ 0.05m =  200 N/m

i.e. if 10N stretches the spring 0.05m then 200N stretches it 1 m

The equation of motion is then:  mx” +  kx  = mg

Or, since m = 1 kg and mg = 10 N:    x” +  200 x  = 10

Then the general solution is:

x = c1 sin (10)t + c2 cos (10) t + 0.05

At t = 0, x = 10 cm (5cm + 5cm) = 0.1m and:

0.10= 0 + c2 cos (0) + 0.05 = c2 + 0.05

Or: c2 = 0.10 – 0.05 = 0.05

Further: x’ = 0 at t = 0 so that (remember your 1st derivatives of sine, cos!)

0 = 10 c1 cos 10(0) – 10 c2sin (10) (0) = 10 c1 cos (10)

or: c1 = 0

Therefore: x = 0.05 cos (10)t + 0.05

The period T = 2 p/w =     2 p/ (10) = p/ 5  s

The frequency f = 1/T = 1/ (p/ 5) =  5/ p

Example (2): Consider the L-C-R series circuit shown.

Here C is a capacitor (which can store charge Q), R is a resistor, and L is an inductance. The effect of inductance is to oppose any change in current and contributes to a voltage drop L (dI/dt). The capacitor meanwhile exhibits a voltage drop of V = C/Q. By Kirchoff’s law (the total p.d. is the sum of the drops around the loop) it must be true that:

L dI/dt + RI  + Q/C  - E = 0

Where E is the electromotive force. But note that the current I is the rate of change of charge Q, so that: I = dQ/dt

Then: dI/dt = d2Q/dt2

So we need to rewrite the differential equation as:

L (d2Q/dt2 )  + R (dQ/dt) + Q/C = E

We know how to form auxiliary equations so we write this one:

LCm2 + (RC) m + 1 = 0

Let a = (LC), b = (RC) and c = 1 then use the quadratic formula:

{-b + Ö (b2 – 4ac)}/ 2a

To obtain: m = {- RC +  Ö (R2 C2  – 4LC)}/ 2LC

The form of the complementary function will then depend on the form of the discriminant, (R2 C2  – 4LC).  The particular integral depends on E, i.e. if E = const. than Q = CE is a particular solution.

Problems for the Math Maven:

1) A 5 lb. weight hangs vertically on a spring whose spring constant k = 10. The weight is pulled 6 inches farther down and released. Find the equation of motion, its period and the frequency. What would the units be for k?

2)     For the circuit shown, consider the special case of NO capacitor. Then the differential equation of interest becomes:  L (dI/dt) + RI = E

Find the solution if we let E = Eo sin wt.

(Hint: take the integrating factor as: exp (ò adt) = exp (at) with a = R/L )