Answer:
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
