Question

I've got parts a-c and understand them. However, I do not understand the rest of the problem and how to solve for the answers in parts d and e. Any explanation would be helpful.

(1 point) A mass of 4 kg stretches a spring 40 cm. The mass is acted on by an external force of F(t) = 97 cos(0.5t) N and moves in a medium that imparts a viscous force of 8 N when the speed of the mass is 16 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 6 cm/s, formulate the initial value problem describing the motion of the mass. Note that the acceleration due to gravity, g, is g = 980 cm/s2 a. Find the spring constant N/cm 98 N s/cm b. Find the damping constant 0.5 c. Set up a differential equation that describes this system. Let y(t) to denote the displacement, in centimeters, of the mass from its equilibrium position, and give your answer in terms of y, y', y". 4y"+0.5y'+98y 97cos (0.5t) d. Find the general solution to your differential equation from the previous part. Use C1 and C2 to denote arbitrary constants. Use t for independent variable to represent the time elapsed in seconds C1e -1/16t)cos(39.595t)+C2e^(-1/16t)sin(39.595t) e. Solve the differential equation with initial conditions describing the motion/the displacement y(t) of the mass from its equilibrium position. y(t) =

Answer 1

d Con sider the given IvP 4 o.5y 98y= 97 Co.St) Curth y(o)=O and yco = 6 The characteristic equaton 4m+0-5m+9 =0 ma -o-5 (o.5-4C4)98) 2C4) 0.5t -1567-75 1567.75 t iC4.949) 16 The complenmentary sofution is CC C (4-949t)+ C2 SinC4-949t)) =e Cob Let tha po1 thclan sou tion be C97 C4 Co.st)) 44D 05D+98 Replaca by -5 C97 Ces Co.5t)) 4 CO-5))+o.5 D+93 T 97 Ce (o.5t) 97+0.5D Multiply and Divide by (97-0.5D)

97-0.5D C97 c(0.5t)) (9)Co.502) ((aT)48-5 D) cas Co-st) AT)-Co.5D) Again replace by - Co.5) 48-5 D CosCo-St) 979 49-5 D Cas Co 5t) Aの+ 0-125 48-5 60.5 SinCo-st)) 94409 9409.125 9409 + 24.25 Sin Co-5t Ip 9409-125 The generol solu tion iA y Ye+ Yp C2 SnC4949t)) (G CasC4949t) : SnCo-st)) t24-25 9409-125 (9409 TCC (49499+sn 9 49) Then 1 re T6 4949an (4.949t)+ 4-949 C2 Cas (4949t)) +el

e) use initial Conditions y(o) =0 yo Ci 9409 OT 9409.125 9409 3 O.99999 9409.125 and yto=6 16 9409 +4.949 C2 6 1G9409-125) 9409 C21 1G (9409-125)x4.949 4.949 C2 =.1997 16 There fore, yCt)- e Co999 c(4949t) +1997Sn (4.949t)) Co.9999 +O.0025773 Sn ((45-0)