Question

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?

Answer 1

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.