A mass m is attached to a spring of constant k placed horizontally on a frictionless table, After extending the mass a distance d from the equilibrium point, it is released.

Question

(a) What is the maximum speed the mass can achieve? 

(b) What is its acceleration at that point? 

(c) What is the distance the mass travels before it comes to a (temporary) stop? 

(d) What will the magnitude of its acceleration be at that point?

Answer 1

A. The law of conservation of energy tells us that the most energy the system can have is equal to the initial potential energy stored in the spring. When velocity is at a maximum, all of the energy becomes kinetic and there is no potential energy so we get the energy balance equation:

1/2 kd²=1/2 mv_max²

so v_max=d*sqrt(k/m)

b. We know maximum values occur when the derivative/slope equals zero. Acceleration is the derivative of velocity, so when velocity is at a maximum, acceleration is 0.

c. It will come to a temporary stop when all of the kinetic energy has been turned back into potential energy, so when displacement is equal to d again (but in the negative direction). Since the block traveled from d to -d, the total distance is 2d

d. Newton's second law states F=ma. The force acting on the mass is the restorative spring force, F=k*d. Therefore a=F/m=k*d/m

I hope that helps, good luck with your physics class, and don't be afraid to message me if you have any more questions!