A container crane lifting a container is hit by a gust of wind as it is lifted at t 0 the horizontal displacement (approximately) of the container , y(t), is given by the equationy''(t) + 2y'(y) + 5y(t) 0,y(0) 0,y'(0) 3a) what do the initial conditions represent physically?b) by letting y e^(λt) find the general solution and then use the initial conditions to find c1 and c2c) the crane driver can not resume work until the oscillations drop to one tenth of their initial maximum amplitude, how long must she wait?
Smoking is done between 250 - 300 degrees FSlow and long.keep a source of moisture in the smoker/grill, such as a shallow pan with apple juice, water, or something similar, and check to make sure it didn't all go up in steamRefill when necessary, and it will help add flavor.Check internal temp of 180 F and clear running juices.
I'd do it around 250 degrees Smoking is typically a low-heat process.
If you have a smoker there is no temp just keep the fire going for at least 6 hours
Initial Conditions: They imply that the container is at equilibrium and the wind is pushing the container by 3 velocity units b) y e^(λt) y' λ e^(λt) y λ? e^(λt) Now, substituting in {1}, λ? e^(λt) + 2 λ e^(λt) + 5 e^(λt) 0 e^(λt) [ λ? + 2 λ + 5] 0 λ? + 2 λ + 5 0.{4} Upon solving {4}, λ [ - B ± √ (B? - 4AC)]/(2A) [-2 ± √(4 - 4 x 1 x 5)]/(2 x 1) -1 ± 2 i General Solution: y (t) e^(α t) [c? cos βt + c? sin βt] y e^(-t) [c? cos 2t + c? sin 2t].{5} y' 2 e^(-t) [- c? sin 2t + c? cos 2t].{6} y (0) 0 c? 0 y'(0) 3 2 c? 3 c? 1.5 Hence, y 1.5 e^(-t) sin 2t{7} Also, y' 3 e^(-t) cos 2t.{8} c) To find the maximum amplitude, y' 0 3 e^(-t) cos 2t 0 cos 2t 0 2t π/2, 3π/2, t (2n-1) π/4 ; n ε N Max amplitude 1.5 e^(-t) sin 2t 1.5 e^(-π/4) sin 2(π/4) a 1.5 e^(-π/4) Hence, Required Amplitude a/10 0.15 e^(-π/4) 1.5 e^(-t) sin 2t 0.15 e^(-π/4) Solve this and you will get the time!!! P.S: Methodology is correct!! Please check the answers
