Skip to main content

PROPAGATION OF NON-UNIFORM STRING


## The string is made up of two different mass density strings
## seperated in the middle. Choose the right one to be ligther
## than the left. So, since velocity is inversely propotional
## to the mass density, than the rigth one is faster also.
## r1=c1*dt/dx and r2=c1*dt/dx
function ynext=propagate_two_parts(ynow,yprev,r1,r2)
## initiallization for, take y as previos one
ynext=ynow;
## we have two parts, hence two loops are needed
## the first part is between x=0 and
## let y(now)/2 for better visiulation.
## floor(x) returns the largest integer not greater than x
## length(x) determines the number of column
## or rows in matrix or vector
##I started the 'for' from x=0 but didn't work
##then started from x=1
## but in this case the
## string_two_parts didn't work
## let x0=i0=2
for i=2:floor(length(ynow)/2)
ynext(i) = 2*(1-r1^2)*ynow(i)-yprev(i)+r1^2*(ynow(i+1)+ynow(i-1));
endfor
## the second part is between x0=(L1+L2) /2 and L1+L2 where
## L1 and L2 are string lenghts respectively
## I started the 2nd for from N2steps
## and finished it by 'ynow' but it
## didn't work. So i tried the following
for i=floor(length(ynow)/2)+1:floor(length(ynow))-1
ynext(i) = 2*(1-r2^2)*ynow(i)-yprev(i)+r2^2*(ynow(i+1)+ynow(i-1));
endfor
endfunction

Comments

Post a Comment

Popular posts from this blog

FACTORIAL

## Function that calculates the factorial of a number ## Usage : f=factorial(n) function f=factorial(n) ## Initialize the output f=1; ## Check whether the input is correct if ( (n<0) || (rem(n,1)~=0) ) printf("n cannot be a negative number. Exiting...\n"); return endif for num=1:n f*=num; endfor endfunction
Post a Comment
Read more

One Dimensional Harmonic Oscillator-Numerov Method

x=[]; h0=1; M=4; N=M+1; x(1)=0; x(N)=x(1)+h0*M; x=x(1):h0:x(N) A=zeros(N); A(1,1)=-2*(5*(x(N)*h0)^2/12+1); A(N,N)=-2*(5*(x(1)*h0)^2/12+1); A(1,2)=1-(x(M)*h0)^2)/12; A(N,M)=1-(x(2)*h0)^2)/12; B=zeros(N); B(1,1)=B(N,N)=-10*(h0^2)/6; B(1,2)=B(N,M)=-(h0^2)/6; for i=2:M B(i,i)=-10*(h0^2)/6; B(i,i-1)=B(i,i+1)=-(h0^2)/6; A(i,i)=-2*(5*(x(N+1-i)*h0)^2/12+1); A(i,i+1)=1-(x(N-i)*h0)^2)/12; A(i,i-1)=1-(x(N+2-i)*h0)^2)/12; end A B
Post a Comment
Read more

NEWTON’S METHOD FOR MINIMUM

##Newton's Method to find ##the minimum of the function F(x)=(x-2)^4-9 ##with the initial guess xmin=1.0 ##Constants and initializations xmin=[]; ##The empty array of x that minimizes the F(x) xmin(1)=1.0; ##Initial value of the xmin Fmin=[]; ##Minimum values of F(x) x=0.0:0.1:4.0; ##Only for plotting purposes F=[]; ##Our examined Function evaluated on x-space Fp=[]; ##First derivative of F(x) wrt x Fpp=[]; ##Second derivative o F(x) wrt x NSteps=50; ##Step number of iteration ##Algorithm for n=1:NSteps Fmin(n)=(xmin(n)-2)^4-9; Fp(n)=4*(xmin(n)-2)^3; Fpp(n)=12*(xmin(n)-2)^2; xmin(n+1)=xmin(n)-Fp(n)/Fpp(n); Fmin(n+1)=(xmin(n+1)-2)^4-9; endfor printf("x*, at which F(x) is minimum, is %1.6f\n",xmin(n+1)) printf("Minimum of F(x) is %1.6f\n",Fmin(n+1)) F=(x-2).^4-9; subplot(2,1,1) plot(x,F) title('Newton^,s Method-F(x) vs x'); xlabel('x'); ylabel('F(x)'); text(2,-7,'\downarrow') text(1.7,-5.6,'(xmin,Fm...
Post a Comment
Read more