A spring with an m -kg mass and a damping constant 5 can be held stretched 2 met

Question

A spring with anm-kg mass and a damping constant 5 can be held stretched 2 meters beyond its natural length by a force of 6 newtons. If the spring is stretched 4 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.
m= A spring with anm-kg mass and a damping constant 5 can be held stretched 2 meters beyond its natural length by a force of 6 newtons. If the spring is stretched 4 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.
m=

Explanation / Answer

ok, this looks like a differential equations question? so:

my''+By'+ky=0 where m is mass B is damping and k is the spring constant

so first find the spring constant with the equaton F=ks====>6=k2

k=3

my''+5y'+3y=0 divide by m y''+(5/m)y'+(3/m)y=0 s^2+(5/m)s+(3/m)=0

quad equation s=(-(5/m)±25/m2-12/m)/2 alright for the system to be critically damped whats inside the sqrt must equal 0 so

25/m2=12/m=====>25=12m======> m=25/12

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