Fill in the blanks. A spring with a mass of 4 kg has a natural length of 0.5 m.

Question

Fill in the blanks.

A spring with a mass of 4 kg has a natural length of 0.5 m. A force of 194.4 N is required to maintain it stretched to a length of 1.1 m. If the spring is stretched to a length of 1.1 m and then released with initial velocity 0, find the position of the mass at time t. From Hooke's Law, the force required to stretch the spring is k(0.6) = 194.4 so k = 194.4/0.6 = 324. Using this value of the spring constant k, together with m = 4, we have 4 d2x / dt2 + = 0 As in the earlier general discussion, the solution of this equation is x(t) = c1cos(9t) + c2sin(9t) We are given the initial condition that x(0) = 0.6. But, from the previous equation, x(0) = c1. Therefore c1 = . Differentiating, we get x't) = -9c1sin(9t) + 9c2cos(9t) Since the initial velocity is given as x'(0) = 0, we have c2 = 0 and so the solution is Since the initial velocity is given as x'(0) = o, we have c2 = 0 and so the solution is x(t) =

Explanation / Answer

1)4x''+324x=0

now if we solve it we get

4m^2+324=0

m=9i and -9i

so x[t]=c1cos9t+c2sin9t

2)

x[0]=c1=0.6

x'[0]=0=9c2

c2=0

so x[t]=0.6cos9t

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