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 0.0240 to 39.1 Hz and the amplitude can be set as high as 1.87 m for low frequencies.
A) What is the maximum velocity to which the pilot is subjected if the frequency is set at 20.9 Hz and the amplitude at 1.94 mm?
B) What is the maximum acceleration to which the pilot is subjected if the frequency is set at 20.9 Hz and the amplitude at 1.94 mm?
x=Asinwt
v=Aw*coswt
a=Aw^2sinwt
a) v=A*2PI*f recommend change Amplitude to m.
b) a=A*(2PIf)^2
To solve these questions, we need to understand the relationship between frequency, amplitude, velocity, and acceleration in simple harmonic motion.
In simple harmonic motion, the displacement of an object is described by an equation of the form:
x(t) = A * sin(2πft)
Where:
- x(t) is the displacement at time t,
- A is the amplitude of the motion,
- f is the frequency of the motion in Hz,
- π is a constant equal to approximately 3.14159.
The velocity of the object is given by the derivative of the displacement equation:
v(t) = A * (2πf) * cos(2πft)
And the acceleration is given by the second derivative:
a(t) = -A * (2πf)^2 * sin(2πft)
Let's now solve the questions.
A) What is the maximum velocity to which the pilot is subjected if the frequency is set at 20.9 Hz and the amplitude at 1.94 mm?
To find the maximum velocity, we need to know the amplitude, which is given as 1.94 mm. We need to convert this to meters by dividing by 1000:
Amplitude (A) = 1.94 mm / 1000 = 0.00194 m
Using the velocity equation, we can now calculate the maximum velocity:
v(max) = A * (2πf)
v(max) = 0.00194 m * (2π * 20.9 Hz)
v(max) ≈ 0.00194 m * 2 * 3.14159 * 20.9 Hz
v(max) ≈ 0.257 m/s
Therefore, the maximum velocity to which the pilot is subjected is approximately 0.257 m/s.
B) What is the maximum acceleration to which the pilot is subjected if the frequency is set at 20.9 Hz and the amplitude at 1.94 mm?
Using the acceleration equation:
a(max) = -A * (2πf)^2
a(max) = -0.00194 m * (2π * 20.9 Hz)^2
a(max) ≈ -0.00194 m * (2 * 3.14159 * 20.9 Hz)^2
a(max) ≈ -0.00194 m * (2 * 3.14159 * 20.9 Hz)^2
a(max) ≈ -0.00194 m * 544.366 Hz^2
a(max) ≈ -1.066 m/s^2
Therefore, the maximum acceleration to which the pilot is subjected is approximately -1.066 m/s^2. Note that the negative sign indicates that the acceleration is in the opposite direction to the displacement.