Question
Solve these please
A spring has natural length 0.75 m and a 5 kg mass. A force of 20 N is needed to keep the spring stretched to a length of 1 m. If the spring is stretched to a length of 1.2 m and then released with velocity 0, find the position of the mass after t seconds. x(t) = A spring with an 8 kg mass is kept stretched 0.3 m beyond its natural length by a force of 24 N . The spring starts at its equilibrium position and is given an initial velocity of 1 m/s. Find the position of the mass at any time t. x(t) = A spring with a mass of 2 kg has damping constant 12, and a force of 8.75 N is required to keep the spring stretched 0.5 m beyond its natural length. The spring is stretched 2 m beyond its natural length and then released with zero velocity . Find the position of the mass at any time t. x(t) =
Explanation / Answer
1) f = kx^2 therfore for f =2 0 and x=1 k= 20
F= 20x^2 a = f/m =4x^2 distance = 0.5 a t^2 = 2x^2 t^2
2)