Why won't this plot?
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Derek Gabaldon
il 22 Set 2019
Modificato: KALYAN ACHARJYA
il 22 Set 2019
I'm new to Matlab and I'm trying to plot the velocity related to the function for a damped harmonic oscillator. I have been able to plot the function itself, but when I try to plot the velocity of said function, I receive this error:
Error using plot
Vectors must be the same length.
Error in damped_oscillator (line 102)
plot(t,v);
I'm not sure how to solve this problem. I've linked the original code. Any help is much appreciated!
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KALYAN ACHARJYA
il 22 Set 2019
Modificato: KALYAN ACHARJYA
il 22 Set 2019
You can make the privious three plot in single description using for loop, define "a" in array. #Recomended. The error arises because of non equal length of t and v, as v result from diff. Hence I have ignored the last element of t to make them both equal. You can opt out the first one also, as per your requirements.
m=20; %Mass
k=30; %Spring constant
c=0.9; %Damping constant
A=1;
a=0;
g=c/(2*m); %Damping coefficient
wn=sqrt(k/m); %Natural frequency
wd=sqrt(((wn)^2)-(g)^2); %Damped natural frequency
t=linspace(0,40,1000); %Time value
x=A*exp(-g*t).*cos(wd*t-a); %x(t)
subplot(4,3,1),plot(t,x)
title('Question 1')
xlabel('Time')
ylabel('Displacement (m)');
grid on;
c=0.9; %Damping constant
A=1;
a=0;
g= c/(2*m); %Damping coefficient
wn=sqrt(k/m); %Natural frequency
wd=sqrt(((wn)^2)-(g)^2); %Damped natural frequency
t=linspace(0,20,1000); %Time value
x=A*exp(-g*t).*cos(wd*t-a); %x(t)
subplot(4,2,1),plot(t,x)
title('Question 2')
xlabel('Time (s)')
ylabel('Displacement (m)')
grid on;
c=0; %Damping constant
A=1;
a=0;
g= c/(2*m); %Damping coefficient
wn=sqrt(k/m); %Natural frequency
wd=sqrt(((wn)^2)-(g)^2); %Damped natural frequency
t=linspace(0,40,1000); %Time value
x=A*exp(-g*t).*cos(wd*t-a); %x(t)
subplot(4,2,2),plot(t,x)
title('Question 3')
xlabel('Time (s)')
ylabel('Displacement (m)')
grid on;
c=1; %Damping constant
A=1;
a=0;
g= c/(2*m); %Damping coefficient
wn=sqrt(k/m); %Natural frequency
wd=sqrt(((wn)^2)-(g)^2); %Damped natural frequency
t=linspace(0,40,1000); %Time value
x=A*exp(-g*t).*cos(wd*t-a); %x(t)
subplot(4,2,3),plot(t,x)
c=5; %Damping constant
A=1;
a=0;
g= c/(2*m); %Damping coefficient
wn=sqrt(k/m); %Natural frequency
wd=sqrt(((wn)^2)-(g)^2); %Damped natural frequency
t=linspace(0,40,1000); %Time value
x=A*exp(-g*t).*cos(wd*t-a); %x(t)
subplot(4,2,4),plot(t,x)
c=15; %Damping constant
A=1;
a=0;
g= c/(2*m); %Damping coefficient
wn=sqrt(k/m); %Natural frequency
wd=sqrt(((wn)^2)-(g)^2); %Damped natural frequency
t=linspace(0,40,1000); %Time value
x=A*exp(-g*t).*cos(wd*t-a); %x(t)
subplot(4,2,5),plot(t,x)
c=50; %Damping constant
A=1;
a=0;
g= c/(2*m); %Damping coefficient
wn=sqrt(k/m); %Natural frequency
wd=sqrt(((wn)^2)-(g)^2); %Damped natural frequency
t=linspace(0,40,1000); %Time value
x=A*exp(-g*t).*cos(wd*t-a); %x(t)
subplot(4,2,6),plot(t,x)
legend('0','1','5','15','50')
t=linspace(0,40,1000); %Time value
x=A*exp(-g.*t).*cos(wd.*t-a); %x(t)
v=diff(x)./diff(t); %Velocity
subplot(4,2,7),plot(t(1:end-1),v);
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Più risposte (1)
MS
il 22 Set 2019
Modificato: MS
il 22 Set 2019
The two vectors have different sizes. Since you are using the diff function. From documentation:
Y = diff(X) calculates differences between adjacent elements of X along the first array dimension whose size does not equal 1:
- If X is a vector of length m, then Y = diff(X) returns a vector of length m-1.
Hence, the t vector has a size of 1 x1000 and the v vector has a size of 1x 999.
if you want, you can update t by removing the first element where the speed is not calculated:
t = t(2:1000)
plot(t,v);
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