**Diagram for Problem 2 of Applications**

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

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 1

^{st}derivatives of sine, cos!)

**10) (0) = 10 c1 cos**

*(***10)**

*(***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 = d

^{2}Q/dt^{2}So we need to rewrite the differential equation as:

L (d

^{2}Q/dt

^{2}) + R (dQ/dt) + Q/C = E

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

LCm

^{2}+ (RC) m + 1 = 0

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

{-b

__+__Ö (b^{2}– 4ac)}/ 2a
To obtain:
m = {- RC

__+__Ö (R^{2 }C^{2 }– 4LC)}/ 2LCThe form of the complementary function will then depend on the form of the discriminant, (R

^{2 }C

^{2 }– 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 = E

_{o}sin wt.
(Hint: take the integrating factor as: exp (ò adt) =
exp (at)
with a
= R/L )

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