A frictionless spring with a 8-kg mass can be held stretched 1.2 meters beyond i
ID: 1947922 • Letter: A
Question
A frictionless spring with a 8-kg mass can be held stretched 1.2 meters beyond its natural length by a force of 10 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 0.5 m/sec, find the position of the mass aftertseconds.
A frictionless spring with a 8-kg mass can be held stretched 1.2 meters beyond its natural length by a force of 10 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 0.5 m/sec, find the position of the mass aftertseconds.A frictionless spring with a 8-kg mass can be held stretched 1.2 meters beyond its natural length by a force of 10 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 0.5 m/sec, find the position of the mass aftertseconds.
Explanation / Answer
8*9.81 + 10 = k*1.2
So, stiffness of spring k = 73.73 N/m
natural frequency = (k/m) = (73.73/8) = 3.036 rad/s
Position x = A Sin(t)
Velocity v = dx/dt = A Cos(t)
At t = 0 s, v = 0.5 m/s. Thus, A = 0.5/3.036 = 0.165 m
Thus, position of mass after t seconds, x = 0.165*Sin(3.036*t)