Answer:
Step-by-step explanation:
Given that,
Spring constant k = 3N/m
Mass attached m=1kg
The motion takes place in a viscous fluid that offers a resistance numerically equal to two times the magnitude of the instantaneous velocity. This implies that the damping coefficient is ζ =2
There is zero external force
I.e F=0
Extension e = 0.4m
Generally, the equation of a mass spring system is give as
mu'' + ζ u' + ku = F(t)
Then, inserting the given datas
u'' + 2u' + 3u = 0
Solving this differential equation using D operator
Then, the characteristics equation is
D² + 2D + 3 =0
Using formula method
D = (—b ± √( b² —4ac) ) /2a
a =1, b = 2 and c =3
D = (—2 ± √(2²—4×1×3)) / (2×1)
D = (—2 ± √(4-12) ) /2
D = (—2 ± √-8 )/2
D = (—2 ± 2√2 •i)/2
D = —1 ± √2 •i
Then, the particular solution
is
Up=Aexp(-t)Cos√2t+Bexp(-t)Sin√2t
No homogeneous solution(Uh) since, F(t) = 0
U(t) = Up + Uh
Therefore,
U(t)= Aexp(-t)Cos√2t+Bexp(-t)Sin√2t
Initial condition given
The initial velocity is 0, i.e U'(0)=0
So, let find U'
U'=-Aexp(-t)Cos√2t — A√2 exp(-t)Sin√2t —Bexp(-t)Sin√2t + B√2exp(-t)Cos√2t
So,
U'(0)=-Aexp(0)Cos√2•0 — A√2 exp(0)Sin√2•0 —Bexp(0)Sin√2•0+ B√2exp(0)Cos√2•0
U'(0)=-Aexp(0)Cos√2•0 — A√2 exp(0)Sin√2•0 —Bexp(0)Sin√2•0+ B√2exp(0)Cos√2•0
U'(0)=-A+B
0 = - A+B
A=B
Also, second condition at t=0, u=4
U(t)= Aexp(-t)Cos√2t+Bexp(-t)Sin√2t
U(0) = Aexp(0)Cos√2•0+Bexp(0)Sin√2•0
U(0) = A
4 = A
Since, A=B
Then, B=4
So, the general solution becomes
U(t)= 4exp(-t)Cos√2t+4exp(-t)Sin√2t
So, this is the position at any time
b. Then, value of U(t) at t→+∞
U(t)= 4exp(-t)Cos√2t+4exp(-t)Sin√2t
U(∞) = 4exp(-∞)Cos∞ + 4exp(-∞)Sin∞
Since exp(-∞) =0
And 1≤CosX≤1, 1≤SinX≤1
U(∞) = 4exp(-∞)Cos∞ + 4exp(-∞)Sin∞
U(∞) = 0 + 0
U(∞) = 0
At infinity, the position is zero.