A spring with an mkg mass and a damping constant 5 kgs can b

A spring with an m-kg mass and a damping constant 5 (kg/s) can be held stretched 1.5 meters beyond its natural length by a force of 7.5 newtons. If the spring is stretched 3 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping. m = 3.125 kg

Solution

Here using hook`s law the damping equation will be described based on values :

m=m, c= 5 and k = 7.5/1.5=5(It is because resulting force is always k times of displacement)

Now as critical damping occurs when c^2-4mk =0

so 5^2-4m(5)= 0

or 25-20m =0

or m= 25/20 = 1.25 kg.

that is the required mass that would produce critical damping.