Question

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. the maximum speed of the object is 1.25 m/s, and its maximum acceleration is 6.89 m/s2. how much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum?

Answer 1

First of all, the maximum speed occurs when the object passes through the
 equilibrium position

The kinetic energy when the object has this max speed is

K= 1/2 * mass * (1.25 m/s)^2

The potential energy in the spring when the speed is equal to zero

U= 1/2 * k * xmax^2

The maximun force of the spring is

mass*acceleration = k*xmax

m * 6.89 m/s2 = k * xmax
xmax = 6.89* m / k

0.5 * m * 1.56  = 0.5 * k * xmax^2

m * 1.56  =  k * (6.89* m / k )^2

1.56 m = 47.47 m^2 / k
m/k = 0.032862

period = 2 *pi*sqrt[m/k]
= 2 pi sqrt [ 0.032862]
= 1.139  s

  A fourth of the period elapses between the instants of max acceleration and maximum speed

= 1/4* period
= 1/4 * 1.139  s = 0.284s