The discussion revolves around a mass-spring system where a 0.824 kg mass attached to a vertical spring with a force constant of 162 N/m oscillates. The original poster seeks to calculate the maximum acceleration of the mass given its maximum speed of 0.372 m/s.
Participants are exploring the dynamics of the mass-spring system, particularly the differences in forces at the extremes of motion. Some guidance has been offered regarding the net forces at the lowest and highest points of oscillation, but there remains a lack of consensus on the implications of these differences for acceleration.
Participants are discussing the implications of the mass oscillating about the equilibrium point and the effects of gravitational force on the spring force at different positions. There is an ongoing examination of the assumptions related to the system's behavior in vertical versus horizontal orientations.
Realize that the mass oscillates about the equilibrium point, where kx0 = mg.Li109 said:I am a bit confused on how to calculate the acceleration of the mass. Are the accelerations at the bottom and the top the same or different? at eq we have kx=mg. when you displace the mass downward and release, ma=kx-mg. when the mass reaches the top, kx+mg=ma. since x is the same in both cases, the two a's need to be different. what is the flaw in my line of thinking?
Doc Al said:Realize that the mass oscillates about the equilibrium point, where kx0 = mg.
At the lowest point, the net force is: k(x0 + A) - mg = kA (upward)
At the highest point, the net force is: k(x0 - A) - mg = -kA (downward)
Right.Li109 said:so this means that the force due to the spring at the top and the bottom are different?
If the system was horizontal, the equilibrium point would be at the unstretched position, x = 0. Thus at one extreme the spring force would be kA and at the other it would be -kA.while if the system was horizontal, it'd be the same at both extremes?