An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.47 m/s, and its maximum acceleration is 6.24 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?

t = 0.37 seconds

Explanation:

t = (1/4)T

Maximum acceleration is;

a_max = Aω²

In simple harmonic motion, we know that v_max = Aω

Thus, a_max = v_max•ω

ω = a_max/v_max

We know that Period is given by;

T = 2π/ω

From initially, t = (1/4)T so, T = 4t

Thus, 4t = 2π/(a_max/v_max)

t = (2π/4)(v_max/a_max)

We are given;

Maximum velocity;v_max = 1.47 m/s

Max acceleration;a_max =6.24 m/s²

Thus,

t = (2π/4)(1.47/6.24)

t = 0.37 seconds

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