Question

The blackbox recorder from an airplane that hit a patch of severe turbulence indicated that the plane moved up and down with an amplitude of 30.0 m and a maximum acceleration of 1.8 g. Assuming the vertical motion was SHM Find the period Find the plane’s maximum vertical speed

Answer 1

T = 8.19 s ,
`v_(max)` = 23 m / s

Step-by-step explanation:

In the simple harmonic motion the equation that describes them is

y = A cos wt

Acceleration can be found by derivatives

a = d²y / dt²

v = dy / dt = - Aw sin wt

a= d²y / dt² = - A w² cos wt

For maximum acceleration cosWT = + -1

`a_(max)` = -A w2

w = RA (
`a_(max)`/ A)

w = RA (1.8 9.8 / 30.0)

w = 0.767 rad / s

The angular velocity is related to the frequency

w = 2π f

f = 1 / T

w = 2π / T

T = 2π / w

T = 2π / 0.767

T = 8.19 s

For maximum speed the sin wt = + -1

`v_(max)` = A w

`v_(max)` = 30.0 0.767

`v_(max)` = 23 m / s