1) A spring hangs vertically. A weight of 10 lbs. is attached and the spring str

Question

1) A spring hangs vertically. A weight of 10 lbs. is attached and the spring stretches 2 inches. The weight is replaced by a 650 lbs. weight and allowed to come to rest in equilibrium. It is then pulled down 6 inches and released with no veolicty. Find the formula for the resulting motion.

2) A 10 kg mass is attached to a spring which i thereby stretched .7 meters from its natural length. The mass is started in motion from the equilibrium position with an initial velocity of 1 m/s in the upward direction. Find the resulting motion if there is a friction coecient of 90.

3) A mass of .25 slug is attached to a spring of force contant k = 1 lbs./ft. The mass is set in motion by initially displacing it 2 ft. in the downward direction and giving it an initial velocity of 2 ft./sec. in the upward direction. Find the subsequent motion of the mass if the friction coecient is  = 1.

Explanation / Answer

1)

first we need to find the spring constant

weight = 10lbs

spring force = kx = k*2

k*2 = 10

k = 5 lb/inch

Now the weight is pulled down by 6 inch,so amplitude of motion is 6inch

general equation of SHM is Asin(wt),but the spring is released at extreme position so equation is

Asin(wt+90) = Acoswt

w = sqrt(k/m) = sqrt(5/10) = 0.7

equation of motion = 6cos(0.7t) (considering the equilibrium position as reference point)

2)

firstly finding the spring constant

mg = kx

10*10 = k*0.7

k = 142.85 n/m

The motion can be described by X(t) = Asin(wt)

velocity = dx/dt

hence v(t) = Awcos(wt) , w = sqrt(k/m) = 3.8

At current position i.e, t = 0 , velocity = Aw

Aw = 1m/s

A = 1/w = 0.26

Hence equation of motion is 0.26sin(wt) (considering equilibrium position as reference and upward as positive)

3)

0.25 slug = 8.04 lbs

w = sqrt(k/m)

= sqrt(1/8.04) = 0.35

equation of motion will be Asin(wt + theta) [since at t=0 it is not at mean position]

Asin(wt+theta) = 2 (t = 0)

Asin(theta) = 2

velocity = dx/dt

Awcos(theta) = 2

dividing them gives tan(theta) = 1 => theta = 135 deg (since voltage is in upward direction and displacement is downward)

Asin(theta) = 2

Asin(135) = 2

A = 2*sqrt(2) = 2.8 ft

equation of motion is 2.8sin(0.35t+135)

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