A mass weighing 4 lb stretches a spring 1.5 in. The mass is given a positive displacement of 2 in from its equilibrium position and released with no initial velocity. Assuming that there is no damping and that the mass is acted on by an external force of 2 cosp3tq lb, formulate the initial value problem describing the motion of the mass.

Explanation:

Given that,

Weight is 4lb

W=4lb

Also extension the weight cause is 1.5in

e=1.5in

1ft=12in

Then, e=1.5in=1.5/12=⅛ft

e=0.125ft

Initial displacement with no velocity is 2in

Then, U(0) = 2in=2/12=1/6ft

Given that

F(t)=2Cos3t lb

No damping system,

Generally, the differential equation governing the spring system is as

mU''+γU′+kU=F(t)

Where

m is the mass attached to spring

γ is the damping constant and it is zero in this case γ=0

k is spring constant

Along with this differential equation we will have the following initial conditions.

U(0)=1/6

U'(0)=0

Then the equation without damping becomes

mU"+ kU= F(t)

Since we are given weight, let calculate the mass

Weight is given as

W=mg, g=32ft/s²

m=W/g

m=4/32

m=⅛ lbs²/ft

To get spring constant k

From hooke's law

F=ke

k=F/e

Where F is the weight

k=4/0.125

k=32lb/ft

Then, the initial value problem becomes

mU′′+ kU= F(t)

⅛U'' + 32U = 2Cos3t

Multiply through by 8

U''+256U=16Cos3t

Initial value are

U(0)= 1/6 and U'(0)=0

Where u is in ft and t is in seconds

correct it is c.) b b