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Chapter 10: Problem 40

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 0.02 to $40.0 \mathrm{Hz}\( and the amplitude can be set as high as \)2 \mathrm{m}$ for low frequencies. What are the maximum velocity and acceleration to which the pilot is subjected if the frequency is set at \(25.0 \mathrm{Hz}\) and the amplitude at \(1.00 \mathrm{mm} ?\)

Short Answer

Expert verified
Answer: The maximum velocity experienced by the pilot is approximately 0.1571 m/s, and the maximum acceleration is approximately 7.8891 m/s².

Step by step solution

01

Identify the given information

From the exercise, we know the frequency is \(f = 25.0 \mathrm{Hz}\) and the amplitude is \(A = 1.00 \mathrm{mm} = 0.001 \mathrm{m}.\)
02

Calculate the angular frequency

Angular frequency, \(\omega\), is related to frequency, \(f\), by the formula: $$\omega = 2\pi f$$ Plugging in the given frequency, we get: $$\omega = 2\pi (25.0 \mathrm{Hz}) = 50 \pi \mathrm{rad/s}$$
03

Calculate the maximum velocity

The maximum velocity, \(v_{max}\), in simple harmonic motion is given by the formula: $$v_{max} = A\omega$$ Using the calculated angular frequency and given amplitude, we can find the maximum velocity: $$v_{max} = (0.001 \mathrm{m})(50 \pi \mathrm{rad/s}) = 0.05 \pi \mathrm{m/s}$$
04

Calculate the maximum acceleration

The maximum acceleration, \(a_{max}\), in simple harmonic motion is given by the formula: $$a_{max} = A\omega^2$$ Using the calculated angular frequency and given amplitude, we can find the maximum acceleration: $$a_{max} = (0.001 \mathrm{m})(50 \pi \mathrm{rad/s})^2 = 0.08 \pi^2 \mathrm{m/s^2}$$
05

Present the results

The maximum velocity to which the pilot is subjected is: $$v_{max} = 0.05 \pi \approx 0.1571 \mathrm{m/s}$$ The maximum acceleration to which the pilot is subjected is: $$a_{max} = 0.08 \pi^2 \approx 7.8891 \mathrm{m/s^2}$$

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