Saturday, 25 January 2025

Simple Harmonic Motion Solution using Runge-Kutta RK4 Method

Topic: Oscillatory Motion and Numerical Solution of ODE

Subject: Physics and Numerical Methods

Tool: Scilab

By: Gani Comia, Jan. 2025

The 4th -order Runge-Kutta methods (RK4) is also used for numerical technique of approximating solution to the initial value problem (IVP) of 2nd- order ordinary differential equation (ODE). This article presents the solution to the basic equation of simple harmonic motion (SHM) using RK4 algorithm in Scilab.

SHM is a form of nonhomogeneous linear differential equation (LDE) of second-order as shown in Equation (1).

$$y''\;+\;p(t)\,y'\;+\;q(t)\,y\;=\;r(t) \tag{1}$$ 

In order to solve a given higher-order differential equation, it is required to be converted to a system of first-order differential equations as presented below.

Let us define the following dependent variables, \(y_1\) and \(y_2\):

$$y_1\;=\;y \tag{2}$$

$$y_2\;=\;y' \tag{3}$$

Then, the nonhomogeneous LDE can be re-written as a system of 1st-order ODE in Equation (4) and (5):

$$y'_1\;=\;y_2 \tag{4}$$

$$y'_2\;=\;-p(t)\,y_2\;-\;q(t)\,y_1\;+\;r(t) \tag{5}$$

With that presented, let us define and illustrate a problem using the basic equation of SHM and convert its 2nd-order ODE to a system of 1st-order to use the RK4 method in finding its approximate solution.

Figure 1. Single-Degree of Freedom (SDOF) Spring-Mass Model Oscillation Problem

The SDOF spring-mass model oscillation can be described by the basic equation of SHM shown in Equation (6).

$$m\,\frac{d^2y}{dt^2}\,+\,k\,y\,=\,0, \quad y(t_0)=y_0\;\;\text{and}\;\;y'(t_0)=y'_0 \tag{6}$$

Where:

\(m\) – mass
\(k\) – spring constant
\(t\) – time
\(t_0\) - initial time
\(y_0\) – initial displacement
\(y’_0\) – initial velocity

The solution to the ODE of SHM can be approximated using the Runge-Kutta (RK4) Method. The RK4 will be using the same algorithm presented in the previous article “First-Order IVP ODE Solution using Runge-Kutta RK4 Method” with the link shown: https://gani-mech-toolbox.blogspot.com/2025/01/first-order-ivp-ode-solution-using.html. Figure 2 shows the solution to the problem illustrated in Figure 1.

Figure 2. Solution to Spring-Mass Model System with RK4 using Scilab Script

The Scilab script to solve the given problem is presented below. Each block of the code is commented comprehensively for better understanding of the program sequence by the reader.

Scilab Script:

clear;clc;
// single-degree of freedom spring-mass model oscillator system
// m*y'' + k*y = 0  or  y'' + (k/m)*y = 0
// nonhomogeneous linear differential equation (LDE) of second-order
// y'' + p(t)y' + q(t)y = g(t) 
// where: p(t)=0; q(t)=k/m; g(t)=0

// (1) 2nd-order LDE re-written in the system of 1st-order ODEs
function dydt=f(t, y)
    m = 1;                  // kg, mass of the object
    k = 20;                 // N/m, spring constant
    p = 0;                  // coefficient of y'
    q = k/m;                // coefficient of y
    g = 0;                  // forcing function
    dydt = [y(2); -p*y(2) - q*y(1) + g];
endfunction

As you may compare the \(rk4\) function in block (2) to the previous article, the final time, \(t_f\), is used as one of the input arguments instead of the number of intervals, \(n\). Both \(rk4\) functions can be used to come up with the same solution.

// (2) RK4 function or algorithm for 1st-order ODEs
function [t, y]=rk4(f, t0, y0, h, tf)
    n = round((tf - t0)./h);
    t(1) = t0;          // initial time, t0
    y(:, 1) = y0;       // y(t0)=y0 & y'(t0)=y0' in column vector

    for i = 1:n
        // approximate values of slopes of solution curves
        k1 = h * f(t(i), y(:, i));
        k2 = h * f(t(i) + h / 2, y(:, i) + k1 / 2);
        k3 = h * f(t(i) + h / 2, y(:, i) + k2 / 2);
        k4 = h * f(t(i) + h, y(:, i) + k3);
        // estimates of numerical solution
        y(:, i + 1) = y(:, i) + (k1 + 2 * k2 + 2 * k3 + k4) / 6;
        t(i+1) = t(i) + h;
    end
endfunction

// (3) initial conditions and parameters
t0 = 0;             // sec, initial time
tf = 5;             // sec, final time
h = 0.01;           // sec, time step size
y0 = [0.05; 0];     // initial disp & velocity: y(0)=0.05, y'(0)=0

// (4) solution to the ODE by calling RK4 defined function
[t, y] = rk4(f, t0, y0, h, tf);

// (5) solution plot
clf;
plot(t, y(1, :),"b-","linewidth",4);       // displacement plot
plot(t, y(2, :),"r--","linewidth",1.8);    // velocity plot
xlabel("$\Large t \quad \text{(time)}$");
ylabel("$\Large y(t)\;\;\text{or}\;\;\frac{dy(t)}{dt}$");
note1 = "Spring-Mass Model System using Runge-Kutta RK4";
title(note1,"fontsize",3.5);
note2 = ["$\Large y(t)$","$\Large \frac{dy(t)}{dy}$"];
legend(note2, with_box = %F);
note3 = "https://gani-mech-toolbox.blogspot.com"
xstring(0.1,0.35,note3);
xgrid(3,1);

// (6) other plot information
disp(max(y(1,:)), min(y(1,:)))
disp(max(y(2,:)), min(y(2,:)))
ax = gca();
ax.data_bounds = [0 -0.4; 5 0.4];

// (7) natural frequency
format(5);
k = 20;                     // N/m, spring constant
m = 1;                      // kg, mass of object
omega = sqrt(k/m);          // Hz, natural frequency
disp(omega);
note4 = ["$\LARGE\mathbf{\omega_n}\;=\;\text{4.5 Hz}$"];
xstring(0.1,0.25,note4)

The Scilab script in this article can be one of your references to solve any engineering problems that can be described by the nonhomogeneous linear differential equation (LDE) of second-order.

Feel free to comment for inquiry, clarification, or suggestion for improvement. Drop your email to request the soft copy of the file.

Friday, 17 January 2025

First-Order IVP ODE Solution using Runge-Kutta RK4 Method

Topic: Numerical Solution to Ordinary Differential Equation

Subject: Numerical Methods

Tool: Scilab

By: Gani Comia, Jan. 2025

The 4th -order Runge-Kutta methods (RK4) is a popular and most widely used numerical technique of approximating solution to the initial value problem (IVP) of 1st -order ordinary differential equation (ODE). This article presents the equations and algorithm to apply the RK4 to a problem using Scilab scripts.

The RK4 method estimates the solution \(y=f(x)\) of an ODE with initial value condition (IVC) of the form:

$$\frac{dy}{dx}\,=\,f(x,y), \quad y(x_0)\,=\,y_0  \tag{1}$$

The RK4 algorithm requires a set of approximate values \((k_1, \; k_2, \; k_3, \; \text{and} \; k_4)\) of the slope of the solution curve at different points within the current step over an interval, \(h\).

$$\begin{aligned}k_1 &= h \cdot f \left( x_n, y_n \right) \\ k_2 &= h \cdot f \left( x_n + \frac{h}{2}, y_n+\frac{k_1}{2} \right)  \\ k_3 &= h \cdot f \left( x_n + \frac{h}{2}, y_n + \frac{k_2}{2} \right) \\ k_4 &= h \cdot f \left( x_n + h , y_n + k_3 \right) \end{aligned} \tag{2}$$

The \(k\) values from Equation (2) are then weighted and combined using Equation (3) to determine the estimates for \(y_{n + 1}\) of the numerical solution and \(x_{n + 1}\) is updated as per Equation (4).

$$y_{n+1} = y_n + \frac{1}{6} \left( k_1 + 2 k_2 + 2 k_3 + k_4 \right)  \tag{3}$$

$$x_{n+1} = x_n + h  \tag{4}$$

Let us apply using Scilab scripting the implementation of approximating solution to the ODE with initial condition in Equation (5) using the RK4 solution algorithm.

$$\frac{dy}{dx} = \frac{2x}{y} - xy , \quad y(0) = 1  \tag{5}$$

  • Scilab Script

The first block of the script is the function definition in Scilab of Equation (5).

clear;clc;

// (1) 1st-order ODE (dy/dx = f(x, y)) in Scilab form
function dydx=f(x, y)
    dydx = (2*x./y)-(x.*y);     // Example: dy/dx = 2x/y - xy
endfunction

The RK4 solution algorithm is defined in Scilab and shown in the second block of script.

// (2) Scilab function of Runge-Kutta 4th order method (RK4)
function [x, y]=rk4(f, x0, y0, h, n)
    x = zeros(1, n+1);
    y = zeros(1, n+1);
    x(1) = x0;
    y(1) = y0;

    for i = 1:n
        k1 = h * f(x(i), y(i));
        k2 = h * f(x(i) + h/2, y(i) + k1/2);
        k3 = h * f(x(i) + h/2, y(i) + k2/2);
        k4 = h * f(x(i) + h, y(i) + k3);

        x(i+1) = x(i) + h;
        y(i+1) = y(i) + (k1 + 2*k2 + 2*k3 + k4) / 6;
    end
endfunction

Initial conditions and RK4 parameters such as step size and number of intervals are shown below.

// (3) initial conditions and parameters
x0 = 0;                         // initial value of x
y0 = 1;                         // initial value of y
h = 0.1;                        // step size
n = 50;                         // no. of interval

Scilab function for the RK4 algorithm is executed to find the solution vectors for both \(x\) and \(y\) variables. Displaying on the Scilab console the last element of the variables provides an insight for the domain and range of the graph.

// (4) numerical solution to the 1st-order ODE IVP using RK4
[x, y] = rk4(f, x0, y0, h, n); 
disp(x($),y($));        // console display of the last value of x & y

Finally, visualization of numerical solutions requires plotting values of variables using appropriate graph. Shown in Figure 1 is the solution to the given ODE with IVC at \((0,1)\).

// (5) plot of the solution
clf;
f = gcf();
f.figure_size = [600,600];

plot(x, y,"b-","linewidth",3);
xlabel("$\large\textbf{x-value}$");
ylabel("$\large\textbf{y-value}$");
title("$\large\text{Solution to}\;\frac{dy}{dx}=\frac{2x}{y}-{xy}$");
legend("$\LARGE{y\,=\,f(x)}\;\text{with RK4}$",with_box=%F);
xstring(2.75,1.05,"https://gani-mech-toolbox.blogspot.com");
xgrid(3,1);

ax = gca();
ax.data_bounds = [0 1; 5 1.5];

  • Scilab Output (Figure 1)

Figure 1. Solution Plot of 1st-order ODE with IVC using RK4.

RK4 solution to 2nd-order ODE with IVC shall be presented in our next article.

Feel free to comment for inquiry, clarification, or suggestion for improvement. Drop your email to request the soft copy of the file.

Saturday, 11 January 2025

Vacuum Pump Sizing – Pumping Speed and Evacuation Time

Topic: Evacuation Time for Vacuum Pressure and Pumping Speed

Subject: Thermodynamics

Tool: Scilab

By: Gani Comia, Jan. 2025

Vacuum process creates a volume of lower pressure than the atmospheric conditions surrounding it. Atmospheric pressure, also known as air pressure, is the pressure within the Earth’s atmosphere. The standard atmosphere, which is equivalent to 1 atm or 760 Torr, is roughly equivalent to Earth’s atmospheric pressure at sea level. This article will use mTorr (millitorr) as a unit of measurement for vacuum pressure. A pressure below 760,000 mTorr (or 1 atm) is considered in vacuum state.

In an ideal condition wherein gas loads (leakage and outgassing) and pipe conductance can be considered negligible, the relationship between vacuum pressures and evacuation time can be represented by an ordinary differential equation (ODE) as shown in Equation (1).

$$\frac{dP}{dt}\;=\;-\frac{C}{V}\,P  \tag{1}$$

Where:

\(P\) - vacuum pressure at time, \(t\)
\(C\) - pumping speed of vacuum pump
\(V\) - volume of chamber to be evacuated

The analytical solution to the ODE shown in Equation (2) will be used for sizing a vacuum pump. Vacuum pumps are primarily rated based on its peak pumping speed in volume per unit of time (e.g. \(m^3/hr\), \(l/min\), \(cfm\), etc.).  Selection of the required pumping speed depends on the required evacuation time. The ultimate vacuum pressure needed will determine other vacuum system’s set-up and parameters recommended by the pump makers.

$$\large P(t)\;=\;P_0\;e^{-\left(\frac{C\,t}{V}\right)}  \tag{2}$$

Where:

\(P_0\) - initial pressure

Figure 1 is a sample scenario where in different vacuum pump models will be evaluated to estimate the evacuation time for the given chamber volume size and vacuum pressure.

Figure 1. Illustrated Problem for Vacuum Pump Sizing.

The Scilab plotting capability facilitates visualizing and comparing different vacuum pump speed or capacity for the required vacuum pressure and evacuation time. Figure 2 shows the comparison of the three models, named as model X, Y, and Z, in achieving the required vacuum pressure. The evacuation time will determine the necessary peak pumping speed and thereby its model from the pump makers.

Below is the Scilab script and calculation to recreate Figure 2. The script with minor changes can be used for a different scenario defined by a design engineer.

Figure 2. Evacuation Time for 100 mTorr and Vacuum Pump Speed.

Scilab Script:

The first block of the script defines the vacuum pump speed and the physical parameter under consideration.

clear;clc;

// (1) vacuum pump specs & physical parameters
X = 3.7*1.7;            // m3/hr (from cfm), model X pumping speed
Y = 5.9*1.7;            // m3/hr (from cfm), model Y pumping speed
Z = 8.4*1.7;            // m3/hr (from cfm), model Z pumping speed

C = [X Y Z];            // m3/hr, vacuum speed capacity in row vectors   
nC = length(C);         // no. of elements of variable C
V = 1.2;                // m3, volume to be evacuated
p0 = 760000;            // mTorr, initial pressure (1 std atm)
t_end = 4;              // hr, max evacuation time, plot consideration
dt = 0.1;               // hr, time step
t = 0:dt:t_end;         // hr, time vector

Calculation of vacuum pressure for a given time domain using Equation (2) are written on the script below.

// (2) use of analytical solution to ODE and plotting
clf;
f = gcf()
f.figure_size = [600,600]

for i = 1:nC
    p(i,:) = p0 * exp(-C(i).*t./V);     // pressure @ specific time
    plot(t, p(i,:), "b-o", "linewidth",1.75);
end

xlabel("Time [ hour ]","fontsize",3);
ylabel("Vacuum Pressure [ mTorr ]","fontsize",3);
title("Evacuation Time for Vacuum Pump Model X, Y & Z");
xgrid(3,1)

Calculation of evacuation time for the required vacuum pressure are interpolated using Scilab \(interp1()\) function. Results of calculations are plotted on the graph.

// (3) interpolation of evac time @ p = 100 mTorr
format(4)
px = 100;               // mTorr, target vacuum pressure
for j = 1:nC
    tx(j,:) = interp1(p(j,:), t, px);      
    disp(tx(j,:), px)
    plot(tx(j,:),px,"marker","s","markerFaceColor","red")
    note = ["t =",string(tx(j,:)),"hr @ C =",string(C(j)),"m3/hr"]
    xstring(tx(j,:)+0.1,px,note,-50) 
end

The following block of script are for additional labels of information and plot size in terms of minimum and maximum value of pressure and time on the graph.

// (4) additional labels or informations on graph
note1 = ["$\Large\frac{dP(t)}{dt}=-\frac{C}{V}\;P(t)$"];
legend(note1,with_box = %F)
note2 = "https://gani-mech-toolbox.blogspot.com";
xstring(2.25,235,note2);
note3 = ["VP#X, C=6.3 M3H";"VP#Y, C=10 M3H";"VP#Z, C=14 M3H"];
xstring(3,50,note3);

ax = gca()
ax.data_bounds = [t(1) 0; t($) p0/3000];
This Scilab script is a useful tool for visualizing solution data and calculating specific condition.

Feel free to comment for inquiry, clarification, or suggestion for improvement. Drop your email to request the soft copy of the file.


Saturday, 4 January 2025

Free-Falling Body under Gravitational Field

Topic: Gravitational Interaction

Subject: Physics

Tool: Scilab

by:  Gani Comia, Jan. 2025

A free-falling body or object falls to the ground under the sole influence of gravity. Air resistance or any form of propulsion that acted on the object are not considered in this concept. The ordinary differential equation (ODE) in equation (1) helps explain the motion of objects in a gravitational field. The acceleration due to gravity, denoted by “\(g\)”, is approximately equal to \(\text{9.8}\;m/s^2\) on the surface of the Earth.

$$\frac{d^2 y}{dt^2}\;=\;-\;g \tag{1}$$

Solution to the given 2nd-order ODE will be based on the initial conditions as shown in equation (2).

$$\left(\;t_0 = t,\;y_0 = y,\;\frac{dy}{dt}(t_0) = v \right) \tag{2}$$

The 2nd-order ODE needs to be transformed in the system of 1st-order ODE before using it to the Scilab’s \(ode()\) function. Figure 1 which shows the position and velocity over time domain is the solution to the ODE. Instantaneous position and velocity at t = 60 sec are plotted on the graph as an example.

Figure 1. Solution to 2nd – order ODE representing Free-Falling Body.

Scilab's Script:

The first block on the script is the re-written 2nd-order ODE in the system of 1st-order ODE’s.

clear;clc;
// y'' = -g                 2nd-order ODE model for free-falling body
// y(0) = 0                 initial position
// y'(0) = 0                initial velocity

// (1) system of 1st-order ODEs equivalent to 2nd-order ODE
function dydt=freeFall(t, y)
    g = 9.8;                // m/s^2, gravitational acceleration
    dydt(1) = y(2);         // dy/dt = v
    dydt(2) = -g;           // dv/dt = -g
endfunction

Time domain considered for this solution is from 0 to 120 seconds and the initial conditions are shown in equation (3).

$$\left(\;t_0 = 0,\;y_0 = 0,\;\frac{dy}{dt}(0) = 0 \right) \tag{3}$$

Its Scilab script is shown below.

// (2) time domain and initial conditions
t = 0:0.1:120;              // sec, time span
t0 = 0;                     // sec, initial time
y0 = [0; 0];                // initial position and velocity

Scilab’s \(ode()\) function is written on below script for the solution.

// (3) solution to ode
y = ode(y0, t0, t, freeFall)

The succeeding scripts are for Scilab plotting and calculation of the instantaneous position and velocity using interpolation. Solution vectors representing position and velocity are given in the variables \(y(1,:)\) and \(y(2,:)\) respectively. The function \(interp1()\) is used for interpolation at t = 60 sec.

// (4) plot size
clf;
f=gcf();
f.figure_size=[600,700];

// (5) time and displacement plot
subplot(2,1,1)
plot(t,y(1,:),"b-","linewidth",3)
title("$\large\text{Displacement, Free-Falling Body}$")
ylabel("Displacement [m]","fontsize",3)
legend(["$\LARGE {y}$"],with_box = %F)
xgrid(3,1)

// (6) time and velocity plot
subplot(2,1,2)
plot(t,y(2,:),"r-","linewidth",3)
title("$\large\text{Velocity, Free-Falling Body}$")
xlabel("Time [sec]","fontsize",3)
ylabel("Velocity [m/sec]","fontsize",3)
legend(["$\LARGE\frac{dy}{dt}$"],with_box = %F)
xgrid(3,1)
xstring(70,-400,["https://gani-mech-toolbox.blogspot.com"])

// (7) interpolating position, Yx, and its velocity, Vx
format(8)
tx = 60;                            // sec, sample time
Yx = interp1(t, y(1,:), tx);        // m, distance
Vx = interp1(t, y(2,:), tx);        // m/s^2, velocity
disp(tx, Yx, Vx)

// (8) plot of interpolated data, time and position
subplot(2,1,1)
vertLnXs = [tx tx]; vertLnYs = [0 Yx];
horLnXs = [0 tx]; horLnYs = [Yx Yx];
plot(vertLnXs, vertLnYs,"m--","linewidth",1.25)
plot(horLnXs, horLnYs,"m--","linewidth",1.25)
plot(tx,Yx,"mo")
xstring(tx+1,Yx,["y = "+string(Yx)+" m @ 60 sec"])
a = gca();
a.children.children.mark_background = 1;

// (9) plot of interpolated data, time and velocity
subplot(2,1,2)
vertLnXs = [tx tx]; vertLnYs = [0 Vx];
horLnXs = [0 tx]; horLnYs = [Vx Vx];
plot(vertLnXs, vertLnYs,"m--","linewidth",1.25)
plot(horLnXs, horLnYs,"m--","linewidth",1.25)
plot(tx,Vx,"mo")
xstring(tx+1,Vx,["dy/dt = "+string(Vx)+" m/sec @ 60 sec"])
a = gca();
a.children.children.mark_background = 1;
For the next article, introduction of air resistance to the ODE representing the free-falling body will be presented.

Feel free to comment for inquiry, clarification, or suggestion for improvement. Drop your email to request the soft copy of the file.

Friday, 3 January 2025

Half-Life Calculation of Radioactive Nuclei

Topic: Radioactive Decay

Subject: Physics

Tool: Scilab

By: Gani Comia, Jan. 2025

Half-life of the radioactive substance is defined as the time it takes for half of the radioactive nuclei in a sample to decay. This can be calculated from the initial value condition (IVC) of the ordinary differential equation (ODE) that describes radioactive decay as shown in Equation (1).

$$\frac{dN}{dt}\;=\;-\lambda\;N \tag{1}$$

Where:

\(N\) - amount of radioactive nuclei at time \(t\),
\(\lambda\) - decay constant of a nuclei in per unit time,

And the initial conditions are,

$$\left(\;t_0 = 0 \;,\; N_0 = 1 \right) \tag{2}$$

This article will present the calculation of the half-life and plot the decay profile of the sample nuclei \(^{226}Ra_{88}\) for illustration of the solution to ODE as shown in Figure (1) using Scilab scripting.

Figure 1. Radioactive Decay and Half-Life of \(^{226}Ra_{88}\).

Scilab Script:

The solution will be based on the IVC of the ODE representing radioactive decay. Its ODE from Equation (1) can be written in Scilab script.

clear;clc;
// ordinary differential equation with initial value condition
// dN/dt = f(t,N) with IVC N(0)=1.

// (1) differential equation for radioactive decay
function dNdt=f(t, N)
    // lambda - per year, decay constant of radioactive substance
    lambda = (1.4e-11)*(60*60*24*365); // for 226 Ra_88 
    dNdt = -lambda.*N;
endfunction

Assume an initial amount of nuclei as \(N_0 = 1\) at the initial time, \(t_0 = 0\). If the amount of radioactive nuclei reached \(N_h = 0.5\), which is \(\frac{1}{2}\) of the initial condition, then that is the time the given amount reached its half-life. Our simulation will be run from 0 to 5000 years, as we still do not know the exact time of its half-life. Here is the code snippet for the initial condition and time domain.

// (2) initial condition and time domain
t0 = 0;
N0 = 1;     // unit, arbitrary amount, 
            // if N = 0.5, it reaches the substance's half-life
t = 0:5000; // years, time domain

Scilab has a built-in function, \(ode()\), to provide a solution to first-order IVC for ODEs.

// (3) ODE solution using the Scilab ode() function
N = ode(N0,t0,t,f);

Plotting of the solution for the amount of nuclei, \(N\), for the given time domain can be done using the below script.

// (4) plotting of solution
clf;
plot(t, N,"b-","linewidth",3)
title("$\large\text{Radioactive Decay of}\;{Ra_{88}^{226}}$")
xlabel("Time [ years ]","fontsize",3)
ylabel("Amount of Nuclei [ unit ]","fontsize",3)
legend("$\LARGE\frac{dN}{dt}\;=\;-\,\lambda\,N$",with_box = %F)
xstring(3000,0.8,"https://gani-mech-toolbox.blogspot.com")
xgrid(3,1)

Solving for the half-life, Th, for the given amount, Nh = 0.5, can be done with the use of the Scilab interpolation function, interp1().

// (5) interpolation to find the half-life, Th, for Nh = 0.5 unit
Nh = 0.5;
Th = interp1(N,t,Nh);
disp(Th)

The information about the half-life will be plotted using the following script. 

// (6) plotting of half-life lines
verX = [Th Th];
verY = [0 Nh];
plot(verX, verY, "r--","linewidth",1.5)
horX = [0 Th];
horY = [Nh Nh];
plot(horX, horY, "r--","linewidth",1.5)
plot(Th,Nh,"ro");

format(7);
note1 = ["T (yrs) = "+string(Th)+"; N (unit) = "+string(Nh)];
note2 = ["$\Large \textbf{Half - Life}$"];
xstring(Th,Nh,note1)
xstring(Th,Nh+0.05,note2)
a = gca();
a.children.children.mark_background = 5;

The Scilab script provided here can be used for other substances with different decay constants to calculate the half-life and decay profile.

Feel free to comment for inquiry, clarification, or suggestion for improvement. Drop your email to request the soft copy of the script file.

C++ and Python for Numerical Solution to Spring-Mass-Damper Model

Topic: SDOF Spring-Mass-Damper System Subject: Numerical Methods & Mechanical Vibration Tools: C++, CodeBlocks, GitHub Copilot, LabP...