A computer to be used in a satellite must be able to withstand accelerations of up to 23 times the acceleration due to gravity. In a test to see if 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 11.6 Hz. What is the minimum amplitude of vibration that must be used in this test?

Ah, the computer is going for a wild joyride in a satellite, eh? Well, let me calculate that for you.

To determine the minimum amplitude of vibration, we need to find the maximum acceleration experienced by the computer.

Now, the maximum acceleration is given by the equation:

amax = ω^2 * A

where ω is the angular frequency and A is the amplitude of vibration.

Given that the acceleration due to gravity is 9.8 m/s^2, we can say that:

23 * 9.8 = (2πf)^2 * A

Plugging in the values, we get:

23 * 9.8 = (2π * 11.6)^2 * A

Solving for A, we get:

A = (23 * 9.8) / (2π * 11.6)^2

Now let me whip out my calculator-bot skills and crunch those numbers for you...

*Calculating intensifies*

After some clowny calculations, the minimum amplitude of vibration required for the test is approximately 0.027 meters. So, strap that computer down and let the dance of high acceleration begin in the satellite!

To find the minimum amplitude of vibration needed in this test, we need to first calculate the maximum acceleration experienced by the computer.

Given:
Acceleration due to gravity (g) = 9.8 m/s^2
Maximum acceleration = 23 times the acceleration due to gravity

We can calculate the maximum acceleration as follows:
Maximum acceleration = 23 * g

Let's substitute the value of g:
Maximum acceleration = 23 * 9.8 m/s^2

Maximum acceleration = 225.4 m/s^2

Now, we can calculate the amplitude of vibration by using the formula for simple harmonic motion:

Amplitude (A) = Maximum acceleration / (2 * π * frequency)^2

Substituting the values we obtained:
A = 225.4 m/s^2 / (2 * π * 11.6 Hz)^2

Calculating further:
A = 225.4 m/s^2 / (2 * 3.14159 * 11.6 Hz)^2

A ≈ 0.1061 meters

Therefore, the minimum amplitude of vibration required for this test is approximately 0.1061 meters.

To determine the minimum amplitude of vibration that must be used in this test, we need to use the formula for the maximum acceleration in simple harmonic motion:

a_max = ω^2 * A

where:
a_max is the maximum acceleration,
ω is the angular frequency (2πf, where f is the frequency in Hz),
A is the amplitude of vibration.

Since we are given the frequency (f = 11.6 Hz) and we want to find the minimum amplitude (A_min), we can rearrange the equation:

A_min = a_max / ω^2

First, let's calculate the angular frequency (ω):

ω = 2πf
= 2π * 11.6 Hz
≈ 72.73 rad/s

Next, we can substitute the given acceleration (23g) into the equation:

a_max = 23 * g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

a_max = 23 * 9.8 m/s^2
= 225.4 m/s^2

Finally, we can calculate the minimum amplitude:

A_min = a_max / ω^2
= 225.4 m/s^2 / (72.73 rad/s)^2
≈ 0.43 m

Therefore, the minimum amplitude of vibration that must be used in this test is approximately 0.43 meters.