Review of Engineering Systems using Ordinary Differential Equations

LO4 Review models of engineering systems using ordinary differential equations
Task 1
Variation of resistance with temperature
a)
dR
dθ = αR
The above equation can be re-written as follows
dθ= dR
αR
Integrating the above equation gives:
∫ d θ= ∫ dR
αR from which θ= 1
α lnR +k
Substituting the initial conditions we have R = Ro when θ=0 gives
0 = 1
α ln Ro +k from which k = - 1
α ln Ro
The solution is therefor of the form, θ= 1
α lnR−¿ 1
α ln Ro = 1
α ¿ ln Ro ¿
Therefore, θ= 1
α ln R
Ro
, αθ=ln R
Ro
R
Ro
=eαθ
R = Roeαθ
b)
with α = 48 × 10-4/°C, Ro = 25, θ=32
R = 25 (e(32 ×48 ×10−4)) = 29.151
Task 2
Equation of motion of an oscillating body:
a)
d2 x
d t 2 + 121x = 0
The above equation is of the form, d2 x
d t 2 + m2x = 0 which represents simple harmonic motion.
Using D operator, the equation can be written as;
(D2 + 121)x = 0
The auxiliary equation is then given by;
M2 + 121 = 0, m2 = -121, m = ± j √ 121 = ±j11

For the complex roots above, the general solution is:
X = e0 ( Acos 11 t+ Bsin 11 t )
X = Acos 11t +Bsin 11t meters
b)
When t = 0, x = 2 and dx
dt =0
dx
dt =¿-11 Asin 11 t+11 Bcos 11t
Therefore, 0 = −11 Asin 0+11 Bcos 0 but sin0 = 0 and cos0 = 1
Thus, 0 = 11 Bcos 0 = 11B, B = 0
Also, cos 11 t+ Bsin 11 t , with t = 0;
Therefore, 2 = Acos11t = A
Therefore the solution is: x = 2cos11t m
Task 3
Newton’s law of cooling:
dθ
dt = -k(θ−θs ¿
Re-arranging the above equation we have,
dθ
θ−θs
= -kdt
Integrating the above equation we obtain, ∫ dθ
θ−θs
= -k∫ dt
From which,
ln(θ−θs) = -kt
ln( θ
θs )= -kt
θ
θs
= e−kt
Therefore, θ=θs e−kt
The initial conditions are: θ=0 at t = 0,
θ0 =θs e0 but e0 = 1
Therefore, θ0 =θs