Need help to be able to script the problem in matlab Second-order differentialek
ID: 3803004 • Letter: N
Question
Need help to be able to script the problem in matlab
Second-order differentialekvationer - damped suspension
we can introduce v (t) = y '(t) and then get the system
(Y '(t) = v (t) (this equation de fi ne v as y'.)
N'(t) = u (t) -G (t) v (t) -h (t) y (t) (note that we replaced with y'' N'and the? 'V,)
More generally, a second order initial value problems
is equivalent to a system of two first order ode with initial values,
We want to describe and calculate the following systems: A basket of some mass m hanging sprung the shock absorption of a wheel following a path with constant speed horizontally. Line height is variable, and is described by p (t). We allow the vertical position of the mass center of gravity at time t is denoted by y = y (t)
Differentialekvationen which defines system
where q (t) = p (t) - y (t) is the distance between the web and the mass center of gravity, and q0 is the length when the system is at rest. The parameters k and m is a spring coefficient and a dampingcoefficient
.
.
We can rewrite the system in two ways
If we describe the path shape (i.e., p (t) using the derivative p '(t) = g (t), and before the v (t) = y' (t), we have the system
where g (t) is a given function.
The file contains a suspd.m the definition of p '(t), and is used by suspensioneq12.m in turn the defined differentialekvationssystemet. The function is defined in the given file suspd.m is a piecewise constant function
Task 4. Simulate system for an appropriate time using ode45. Note that you need to give a vector of initial values consisting of three components, [v (0), y (0), p (0)]. A suitable choice in this case is [0,0,1]. Plotting the basket and the wheel's height, y and p in the same coordinate system. (If t is time vector and x is the calculated solution from ode45, so can draw it with plot (t, x (:, 1) T, x (:, 3), or with plot (t, x) as one also get to the basket vertical speed.) Try some different values of dampingcoefficient (change in suspensioneq.m) and note the differences in the behavior of the system, especially with regard to fluctuations. How does a change of f dampingcoefficient system behavior?
If you have time, please also experimenting with the other parameters with suspd to change banprofile.
Explanation / Answer
public category MyBinarySearch begin = 0;
int finish = inputArr.length - 1;
whereas (start <= end) middle = (start + end) / 2;
if (key == inputArr[mid]) come back mid;
}
if (key < inputArr[mid]) {
finish = middle - 1;
} else begin = middle + 1;
}
}
return -1;
}
public static void main(String[] args) ;
System.out.println("Key 14's position: "+mbs.binarySearch(arr, 14));
int[] arr1 = ;
System.out.println("Key 432's position: "+mbs.binarySearch(arr1, 432));
}
}public category MyBinarySearch begin = 0;
int finish = inputArr.length - 1;
whereas (start <= end) middle = (start + end) / 2;
if (key == inputArr[mid]) come back mid;
}
if (key < inputArr[mid]) {
finish = middle - 1;
} else begin = middle + 1;
}
}
return -1;
}
public static void main(String[] args) ;
System.out.println("Key 14's position: "+mbs.binarySearch(arr, 14));
int[] arr1 = ;
System.out.println("Key 432's position: "+mbs.binarySearch(arr1, 432));
}
}public category MyBinarySearch begin = 0;
int finish = inputArr.length - 1;
whereas (start <= end) middle = (start + end) / 2;
if (key == inputArr[mid]) come back mid;
}
if (key < inputArr[mid]) {
finish = middle - 1;
} else begin = middle + 1;
}
}
return -1;
}
public static void main(String[] args) ;
System.out.println("Key 14's position: "+mbs.binarySearch(arr, 14));
int[] arr1 = ;
System.out.println("Key 432's position: "+mbs.binarySearch(arr1, 432));
}
}public category MyBinarySearch begin = 0;
int finish = inputArr.length - 1;
whereas (start <= end) middle = (start + end) / 2;
if (key == inputArr[mid]) come back mid;
}
if (key < inputArr[mid]) {
finish = middle - 1;
} else begin = middle + 1;
}
}
return -1;
}
public static void main(String[] args) ;
System.out.println("Key 14's position: "+mbs.binarySearch(arr, 14));
int[] arr1 = ;
System.out.println("Key 432's position: "+mbs.binarySearch(arr1, 432));
}
}public category MyBinarySearch begin = 0;
int finish = inputArr.length - 1;
whereas (start <= end) middle = (start + end) / 2;
if (key == inputArr[mid]) come back mid;
}
if (key < inputArr[mid]) {
finish = middle - 1;
} else begin = middle + 1;
}
}
return -1;
}
public static void main(String[] args) ;
System.out.println("Key 14's position: "+mbs.binarySearch(arr, 14));
int[] arr1 = ;
System.out.println("Key 432's position: "+mbs.binarySearch(arr1, 432));
}
}public category MyBinarySearch begin = 0;
int finish = inputArr.length - 1;
whereas (start <= end) middle = (start + end) / 2;
if (key == inputArr[mid]) come back mid;
}
if (key < inputArr[mid]) {
finish = middle - 1;
} else begin = middle + 1;
}
}
return -1;
}
public static void main(String[] args) ;
System.out.println("Key 14's position: "+mbs.binarySearch(arr, 14));
int[] arr1 = ;
System.out.println("Key 432's position: "+mbs.binarySearch(arr1, 432));
}
}