1.A frictionless spring with a 8 slugs mass can be held stretched 1.4 feet beyon
ID: 1836852 • Letter: 1
Question
1.A frictionless spring with a 8 slugs mass can be held stretched 1.4 feet beyond its natural length by a force of 60 lb. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 ft/sec, find the position of the mass after t seconds.
2.A spring with a 3 slugs mass and a damping constant c=11 can be held stretched 0.5 feet beyond its natural length by a force of 1 lb. Suppose the spring is stretched 1 feet beyond its natural length and then released with zero velocity. In the notation of this text, what is the value c^24mk? Find the position of the mass after t seconds. Your answer should be a function of the variable t of the form c1e^(alpha t)+c2e^(beta t) where
alpha = (the larger of the two)
beta= (the smaller of the two)
c1=
c2=
3. A spring with a 5 slugs mass and a damping constant c=2 can be held stretched 2 feet beyond its natural length by a force of 4 lb. Suppose the spring is stretched 4 feet beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c^24mk? Find the position of the mass after t seconds. Your answer should be a function of the variable t with the general form c1e^(alpha t)cos(beta t)+c2e^(gamma t) sin(delta t) where alpha=
beta=
gamma=
delta=
c1=
c2=
Explanation / Answer
1.
1 slug =14.5939 kg
Mass M=8*14.5939 =116.75 kg
1 Lbf =4.44822 N
F=60*4.44822 =266.9 N
1 feet =0.3048 m
x=1.4*0.3048 =0.4267 m
Spring constant
K=F/X =266.9/0.4267 =625.5 N/m
Angular velocity
W=sqrt[K/m] =sqrt[625.5/116.75]
W=2.3 rad/s
The equation for displacement is given by
X(t)=ASin(Wt)
=>V(t)=AWCos(Wt)
at t=0=>V=2 ft/sec =2*0.3048 =0.61 m/s
=>0.61=A*2.3
A=0.264 m
X(t)=(0.264 m)Sin(2.3t)
If answer was need in feet
X(t) =(0.866 ft)Sin(2.3t)
Spring Constant
K=F/x =60/1.4 =42.86 lb/ft
angular velocity
W=sqrt[K/m] =[42.86 Lb/ft/8 Lb-s2/ft] =5.357 rad/s
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