A spring-mass system has a spring constant of 3 N/m. A mass of 1 kg is attached to the spring, and the motion takes place in a viscous fluid that offers a resistance numerically equal to two times the magnitude of the instantaneous velocity. There is zero external force. If the mass is pulled so that the spring is stretched 0.4 m and released (with initial velocity of zero), find the position, u, of the mass at any time t. What is the position, u, as t→+[infinity]

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.

convex lenses are thicker at the middle. rays of light that pass through the lens are brought closer together (they converge). a convex lens is a converging lens. when parallel rays of light pass through a convex lens the refracted rays converge at one point called the principal focus.

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more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .