MATLAB: Error: x(2): out of bound 1

errormatlab functionout of bound

I have code of the following. I did everything I could do, but couldnt overcome this error.
When I execute the below function I get the error
function [disp_dot]=stage2(x,t)%Stage_2 calculation for displacement and pressure
	m=23.5;						     %Stage_2:From bottom choked till bottom is exhausted; top is exhausted
	ODB=0.122;
	IDB=0.04625;
	g=9.8;
	ODT=0.104;
	IDT=0.042;
	Patm=100000;
	dt=0.0001;						%time increment value
	L=0.11;							%This is the length where adiabatic condition starts in top chamber; the length of the leg of the air distributor
	AT = pi*(ODT^2-IDT^2)/4;
	AB = pi*(ODB^2-IDB^2)/4;
	PT=Patm;
	ODSBD=0.1285;
	IDSBD=0.122;
	ASBD=pi*(ODSBD^2-IDSBD^2)/4;
	ASB=0.00451183;
	AST=0.001267892;
	CFM=840;                %air inflow
	CFMm=CFM*(1/2118.88);
	v=CFMm/AB;
	v=5;
	PST=Patm;
	PSB=Patm;
	PT=Patm;
	LB=0.0259;	%This is the length where bottom chamber becomes adiabatic during upward displacement of the piston
	PBinitial=((PT*AT)+(m*g)+(PST*AST)-(PSB*ASB))/AB;
	KB=PBinitial*(AB*LB)^1.4;
	disp_dot(1) = x(2); 
    disp_dot(2) = (1/m)*((KB*AB/(0.000058402+((ASBD+AB)*x(1))^1.4)-(PT*AT)-(m*g)+(PSB*ASB)-(PST*AST)); %Constant value is from bottom channel 4 in excel
end

Best Answer

Issue 1:
When you defined the function as 
[disp_dot]=stage2(x,t)
There is not role of t function input data t in the function.
Issue 2:
In the following one parenthesis bracket required at last.
disp_dot(2)=(1/m)*((KB*AB/(0.000058402+((ASBD+AB)*x(1))^1.4)-(PT*AT)-(m*g)+(PSB*ASB)-(PST*AST)));
Now I have defined the function as 
function [disp_dot]=stage2(x)%Stage_2 calculation for displacement and pressure
	m=23.5;						     %Stage_2:From bottom choked till bottom is exhausted; top is exhausted
	ODB=0.122;
	IDB=0.04625;
	g=9.8;
	ODT=0.104;
	IDT=0.042;
	Patm=100000;
	dt=0.0001;						%time increment value
	L=0.11;							%This is the length where adiabatic condition starts in top chamber; the length of the leg of the air distributor
	AT=pi*(ODT^2-IDT^2)/4;
	AB=pi*(ODB^2-IDB^2)/4;
	PT=Patm;
	ODSBD=0.1285;
	IDSBD=0.122;
	ASBD=pi*(ODSBD^2-IDSBD^2)/4;
	ASB=0.00451183;
	AST=0.001267892;
	CFM=840;                %air inflow
	CFMm=CFM*(1/2118.88);
	v=CFMm/AB;
	v=5;
	PST=Patm;
	PSB=Patm;
	PT=Patm;
	LB=0.0259;	%This is the length where bottom chamber becomes adiabatic during upward displacement of the piston
	PBinitial=((PT*AT)+(m*g)+(PST*AST)-(PSB*ASB))/AB;
	KB=PBinitial*(AB*LB)^1.4;
	disp_dot(1)=x(2); 
    disp_dot(2)=(1/m)*((KB*AB/(0.000058402+((ASBD+AB)*x(1))^1.4)-(PT*AT)-(m*g)+(PSB*ASB)-(PST*AST))); %Constant value is from bottom channel 4 in excel
end
When I ran the code here is the output
>> y1=[1,2]
y1 =
     1     2
>> y=stage2(y1)
y =
    2.0000  -26.1197
Please note in the fuction code x define as 2 length vector, which require x(1) and x(2), that why I have passed the y=[valuue1 value 2] to function
Hope it helps!
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