Y2 Computational
Physics Laboratory
Solutions to Ex. 2
Numerical methods: from a skydiver to
radioactive decay
Problem 1: Free fall motion with air resistance
1. You should include a printout of your code, and a brief description of it. Detail
should be sufficient to allow the marker to easily follow, and understand, the logic
and function of the code. The code itself should be well commented. [2 marks]
Again, the usual comments apply for marking. Explanations should be clear, and easy
to follow; as should the code, which should be well commented.
Here is an example of code to solve the first part.
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
// Declarations
int i; // Integer to loop over time
float alpha,
k,
h,
t,
v,
beta,
va;
// Drag coefficient
// = alpha / mass diver
// Step size for Euler
// Time
// Velocity
// Constant for analytical prediction (term. vel.)
// Analytical solution
const float g=9.81, // Accn. due to gravity
m=100.0; // Mass of diver
// Input values for drag, and step size:
cout << "Input alpha >> ";
cin >> alpha;
cout << "Input h >> ";
cin >> h;
k=alpha/m;
// Calculate k
beta=sqrt(g/k); // Calculate beta
// Into the main loop, setting t=0, v=0 first:
t=0.0;
v=0.0;
for (i=0; i <= 25; i++)
{
t += h;
v += h*(g-k*(v*v));
// Increment t
// Euler estimate
// Analytical value:
va=beta*(1.0-exp(-2.0*beta*k*t))/(1.0+exp(-2.0*beta*k*t));
// Output on the fly
cout << t << " " << v << " " << va << endl;
}
}
2. You should include output of the estimated velocity, and the actual velocity (from
Equation [5]) as a function of time, for at least one value of h. There should also
be a brief discussion of the impact of the choice of h on the accuracy of Euler’s
method. [2 marks]
Here is some sample output from the code, with  = 0.4 and h = 1.0. Give [1] mark
for sensible-looking output.
Input alpha >> 0.4
Input h >> 1.0
1
9.81
9.68367
2
19.2351
18.6541
3
27.5651
26.3937
4
34.3358
32.6724
5
39.43
37.5162
6
43.0211
41.1102
7
45.4278
43.7003
8
46.9831
45.5281
9
47.9634
46.7988
10
48.5715
47.6732
11
48.9447
48.2706
12
49.1724
48.6767
13
49.3107
48.9518
14
49.3945
49.1379
15
49.4452
49.2634
16
49.4759
49.3481
17
49.4944
49.4051
18
49.5056
49.4436
19
49.5124
49.4695
20
49.5165
49.4869
21
49.519
49.4986
22
49.5205
49.5065
23
49.5214
49.5118
24
49.5219
49.5154
25
49.5222
49.5178
26
49.5224
49.5194
Press Enter to return to Quincy...
The discussion over the choice of h should talk about the curvature of the function.
The basic Euler method only works well provided the derivative does not change
significantly over each step of h; which means the step should be kept small! The
more mathematically inclined students might go as far as to derive the expression for
the derivative, which is:
d 2v
dv
 2kv .
2
dt
dt
This approaches zero as v approaches the terminal velocity. It is very small; so Euler
works quite well. (This expression will come in handy later, in Problem 3.)
Award [1] mark for a reasonable, qualitative explanation.
3. You should solve Equation [2] analytically, and include a clearly laid out proof in
your book. [1 mark] You will need to make use of the integral identity:
 v
dv
1

ln
2
   v 2   v .
2
This is a straightforward proof; award 0.5 marks for any attempt.
What is the physical significance of  [0.5 marks]?
This is the terminal velocity of the diver.
Problem 2: Opening the parachute
1. You should include a brief discussion of the effect of h on the results. (Hint: think
about the derivative of the function.) Also describe what you found with your
initial choice of h; and give the values that in the end gave sensible results, saying
why they did so. [1.5 marks]
The modification to the code is very straightforward. If a value of h ~ 1 is still used,
the results become non-physical very quickly. In this new regime, the derivative
changes much more rapidly, i.e., our diver is rapidly decelerated. So a much smaller
value for h is now needed to overcome the numerical shortcomings of the Euler
method. The student must reduce h until the results look sensible; something like h =
0.01 will work.
Problem 3: Radioactive Decay
1. You should include a printout of your code, and a brief description of it. Detail
should be sufficient to allow the marker to easily follow, and understand, the logic
and function of the code. The code itself should be well commented. [1 mark]
Here is some example code:
//
// Radioactive decay, 24/08/05 (WJC)
//
//
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
// Declarations
int i; // Integer to loop over time
float lambda,
h,
t,
n,
na;
// Decay coefficient
// Step size for Euler
// Time
// Estimated number of nuclei
// Analytical solution
// Input values for lambda, and step size:
cout << "Input lambda >> ";
cin >> lambda;
cout << "Input h >> ";
cin >> h;
// Into the main loop, setting t=0, n=1000 first:
t=0.0;
n=1000.0;
for (i=0; i <= 25; i++)
{
t += h;
n += -lambda*h*n;
// Increment t
// Euler estimate
// Analytical value:
na=1000.0*exp(-lambda*t);
// Output on the fly
cout << t << " " << n << " " << na << endl;
}
}
2. You should include output data (Euler and analytical prediction) from your code,
and explain why there is a difference in accuracy compared to the diver problem.
[1 mark]
Here is example output for the required input parameters
Input lambda >> 0.5
Input h >> 1.0
1
500
606.531
2
250
367.879
3
125
223.13
4
62.5
135.335
5
31.25
82.085
6
15.625
49.7871
7
7.8125
30.1974
8
3.90625
18.3156
9
1.95313
11.109
10
0.976563
6.73795
11
0.488281
4.08677
12
0.244141
2.47875
13
0.12207
1.50344
14
0.0610352
0.911882
15
0.0305176
0.553084
16
0.0152588
0.335463
17
0.00762939
0.203468
18
0.0038147
0.12341
19
0.00190735
0.0748518
20
0.000953674
0.0453999
21
0.000476837
0.0275364
22
0.000238419
0.0167017
23
0.000119209
0.0101301
24
5.96046e-005
0.00614421
25
2.98023e-005
0.00372665
26
1.49012e-005
0.00226033
Press Enter to return to Quincy...
This does not do any where near as well as for the diver problem; disagreement
between the Euler and analytical values is quite marked. Again, this is down to the
derivative of the function. Here:
d 2N
dN
 
 2 N .
2
dt
dt
This is big for small t, when N is large; so Euler does not cope too well! This is to be
compared with the equation derived for the derivative of dv / dt above.