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

Answer:

T = 8.19 s ,  $v_{max}$  = 23 m / s

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