Recent content by sarahqwert

Sorry that was supposed to be w(x,t) not u(x,t). So: Xn=C*cos((2n+1)/2)x Tn=D*cos((2n+1)/2)t)+(B*sin((2n+1)/2)t) w(x,t) = Ʃ Xn * Tn w(x,t) = Ʃcos((2n+1)/2)x)*((D*cos((2n+1)/2)t)+(B*sin((2n+1)/2)t)) so w(x,0)=Ʃcos((2n+1)/2)x)*D and w(x,0) = cos(5/2)x

Thank you for your help, I m almost done solving this problem and got: u(x,0)= Ʃ (cos((2n+1)/2)x)*Dn = x sigma from 0 to infinity and Dn is a constant. How do I find Dn?

Homework Statement utt = uxx -(25/4)cos((5/2)x) ux(0,t) =1 u(pi,t)= pi u(x,0)=x ut(x,0)=0 Homework Equations u(x,t)=v(x) + w(x,t) The Attempt at a Solution This is what I did so far: u(x,t)=v(x) + w(x,t) u(x,0) = v(x) +w(x,0) when t is large: vxx - (25/4)cos((5/2)x) = 0 vx =...

Homework Statement a converging diverging nozzle is operating at sea level, it will be mounted on a glider that flies 26.8m/s. The entrance area is 1 m^2. Homework Equations what is the stagnation pressure and mach number of flow entering the nozzle? what should be throat and exit...