A spring is oscillating so that its length is a sinusoidal function of time. Its length varies from a minimum of 10 cm to a maximum of 14 cm. At t=0 seconds, the length of the spring was 12 cm, and it was decreasing in length. It then reached a minimum length at time t = 1.2 seconds. Between time t=0 and t=8 seconds, how much of the time was the spring longer than 13.5 cm?

Let the x (t) represent the motion of the spring as a function of time, t.

The length of the oscillating spring varies from a minimum of 10 cm to a maximum of 14 cm.

Therefore its amplitude is A = (14 - 10) / 2 = 2.

When t = 0 s, x = 12 cm.

Therefore the function is of the form

x (t) = 2 sin (bt) + 12

At t=0, x (t) is decreasing, and it reaches its minimum value when t = 1.2 s.

Therefore, a quarter of the period is 1.2 s.

The period is given by

T/4 = 1.2

T = 4.8 s

That is,

b = (2π) / T = (2π) / 4.8 = π/2.4 = 1.309

The function is

x (t) = 2 sin (1.309t) + 12

A plot of x (t) is shown below.

When x (t) = 13.5, obtain

2 sin (1.309t) + 12 = 13.5

sin (1.309t) = (13.5 - 12) / 2 = 0.75

1.309t = sin⁻¹ 0.75 = 0.8481 or π - 0.8481

t = 0.8481/1.309 or t = (π - 0.8481) / 1.309

= 0.649 or 1.751

The difference in t is 1.751 - 0.649 = 1.1026.

This difference occurs twice between t=0 and t=8 s.

Therefore the spring length is greater than 13.5 cm for 2*1.1026 = 2.205 s.

Answer:

Between t=0 and t=8, the spring is longer than 13.5 cm for 2.205 s.