Question

A certain mass-spring system oscillates with an amplitude of 5mm when the forcing frequency is 20 Hz, and with an amplitude of 1mm when the forcing frequency is 40 Hz. Estimate the natural frequency of the system.

Answer 1

The natural frequency of the mass-spring system can be estimated using the ratio of amplitudes and frequencies.

Step-by-step explanation:

To estimate the natural frequency of the mass-spring system, we can use the relationship between amplitude and frequency. The ratio of the amplitudes is equal to the ratio of the frequencies. In this case, the first amplitude of 5mm corresponds to a frequency of 20 Hz, and the second amplitude of 1mm corresponds to a frequency of 40 Hz.

Using the ratio, we can set up the equation: 5mm/1mm = 20 Hz/fo, where fo is the natural frequency. Solving for fo, we find that the natural frequency of the system is 100 Hz.

Answer 2

The estimated natural frequency of the system is 5.15 Hz.

Step-by-step explanation:

The quality factor Q is a measure of the damping of the system. A higher Q value indicates less damping. The natural frequency fn is the frequency at which the system oscillates with maximum amplitude.

In this case, we can estimate the natural frequency by using the following equation:

fn = f1 / sqrt(1 + Q^2)

where:

f1 is the forcing frequency at which the amplitude is maximum (20 Hz in this case)

Q is the quality factor

We can calculate the quality factor using the following equation:

Q = (f2^2 - f1^2) / (2 * f1 * f2) * (A1 / A2)

where:

f2 is the forcing frequency at which the amplitude is minimum (40 Hz in this case)

A1 is the amplitude at f1 (5 mm in this case)

A2 is the amplitude at f2 (1 mm in this case)

Plugging in the values, we get:

Q = (40^2 - 20^2) / (2 * 20 * 40) * (5 / 1) = 2.5

Plugging this value of Q into the first equation, we get:

fn = 20 / sqrt(1 + 2.5^2) = 5.15 Hz

Therefore, the estimated natural frequency of the system is 5.15 Hz.