A spring hanging from the ceiling is 5 meters long when there is no weight on it

Question

A spring hanging from the ceiling is 5 meters long when there is no weight on it. A body of mass 2 kilograms is attached to the spring and at equilibrium hangs a distance of 7.45 meters from the ceiling.

(a) Assuming that acceleration due to gravity is 9.8 meters/(second)^2, and that air friction is negligible, give the distance y(t) meters of the body from the ceiling "t" seconds after it is pulled to a distance of 8.45 meters below the ceiling and then released.

(b) Now assume that air friction is not negligible and is given by an opposing force of c kg/sec · y(t)

What is the minimum value of "c" such that there will only be a single time "t1" after the body is released when y(t1) = 5 meters?

Explanation / Answer

a)

m = mass attached to spring = 2 kg

k = spring constant

x = stretch in spring while balacing the weight attached = 7.45 - 5 = 2.45 m

using equilibrium of force

mg = kx

2 x 9.8 = k (2.45)

k = 8 N/m

w = angular frequency of oscillation = sqrt(k/m) = sqrt(8/2) = 2 rad/s

A = amplitude = 8.45 - 5 = 3.45 m

equation for displacement is given as

y(t) = A Coswt

y(t) = 3.45 Cos(2t)

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