A computer to be used in a satellite must be able to withstand accelerations of up to 22 times the acceleration due to gravity. In a test to see whether it meets this specification, the computer is bolted to a frame that is vibrated back and forth in simple harmonic motion at a frequency of 9.24Hz. What is the minimum amplitude of vibration that must be used in this test?
To determine the minimum amplitude of vibration that must be used in this test, we need to consider the relationship between acceleration, frequency, and amplitude in simple harmonic motion.
The acceleration in simple harmonic motion can be expressed as:
a = ω^2 * x
Where 'a' is the acceleration, ω is the angular frequency (2πf, where f is the frequency), and 'x' is the amplitude.
Given that the computer must withstand accelerations of up to 22 times the acceleration due to gravity, we can express this as:
a_satellite = 22 * g
Where 'g' is the acceleration due to gravity (approximately 9.8 m/s^2).
To determine the minimum amplitude, we need to find the maximum acceleration:
a_satellite = ω^2 * x_max
Plugging in the values we have:
22 * g = (2πf)^2 * x_max
Substituting the frequency given (9.24 Hz) and solving for x_max:
x_max = (22 * g) / (4π^2 * f^2)
Now, we can substitute the known values and calculate the minimum amplitude:
x_max = (22 * 9.8) / (4π^2 * 9.24^2)
Using a calculator, the minimum amplitude (x_max) is approximately 0.048 meters or 4.8 cm.
Therefore, the minimum amplitude of vibration that must be used in this test is approximately 4.8 cm.