In an aviation test lab, pilots are subjected to vertical oscillations on a shaking rig to see how well they can recognize objects in times of severe airplane vibration. The frequency can be varied from 0.0190 to 38.3 Hz and the amplitude can be set as high as 2.08 m for low frequencies. What is the maximum velocity to which the pilot is subjected if the frequency is set at 22.0 Hz and the amplitude at 1.71 mm?
1.71 mm = 1.71*10^-3 m
y = 1.71*10^-3 * sin (44 pi t)
v = 1.71*10^-3 *44 pi *cos(44 pi t)
max v = 44 pi * .00171 meters/second
To find the maximum velocity to which the pilot is subjected, we can use the formula for the maximum velocity of a harmonic oscillator:
v_max = 2πfA
where:
v_max is the maximum velocity
f is the frequency
A is the amplitude
Given:
f = 22.0 Hz (frequency)
A = 1.71 mm (amplitude)
First, we need to convert the amplitude from millimeters to meters:
A = 1.71 mm = 1.71 x 10^(-3) m
Now, we can substitute the values into the formula:
v_max = 2π(22.0 Hz)(1.71 x 10^(-3) m)
Calculating this equation will give us the maximum velocity to which the pilot is subjected:
v_max ≈ 0.237 m/s (rounded to 3 significant figures)
Therefore, the maximum velocity to which the pilot is subjected is approximately 0.237 m/s.