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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