# Maximum speed in a spring-mass system

I am studying energy right now and I can use only gravitational potential energy, elastic energy and kinetic energy to solve some problems. My doubt is how can I prove that the maximum speed in a mass hanging on a spring is reached at the middle of the elongation of the spring.

The exercise shows the following situation: A block with mass $$m$$ is attached to an ideal spring (no mass) of an elastic constant $$k$$. The block is released when the spring is on his natural state. Consider no friction and that the system is conservative.

The question ask for the maximum speed of the block using energy only. I solved it already (with $$m=0.1$$ $$[kg]$$ and $$k=10\left[\frac{N}{m}\right ]$$ and assuming that the point $$0[m]$$ is when the spring reaches it maximum elongation and the block is released from a height of $$h\;[m])$$. Then, I got that the maximum speed is $$1[\frac{m}{s}]$$, but I assumed that the maximum speed is reached at the middle, so I want to know why this is true in these ideal conditions.

## 2 Answers

The total energy you have is Kinetic + Potential. And for a spring system, the potential energy is minimum when the spring is not stretched or compressed. But since energy is conserved, when potential energy is zero, all the energy in the system is present as kinetic energy. So kinetic energy is maximum when the displacement from natural length of the spring is zero.

• I thought that the total energy is Kinetic, potential and elastic. – Rodrigo Pizarro Jan 16 '20 at 3:54
• The elastic energy is what I considered as potential energy. – Superfast Jellyfish Jan 16 '20 at 3:55

Maximum speed occurs at the mean position of spring mass system as Diffrentiation of velocity, acceleration is 0 there.