Springs are available in stiffness values of 10,100 , and 1000 N / m. Design a spring system us

Springs are available in stiffness values of 10,100 , and...

Springs are available in stiffness values of 10,100 , and $1000 \mathrm{~N} / \mathrm{m}$. Design a spring system using these values only, so that a $100-\mathrm{kg}$ mass is connected to ground with frequency of about $1.5 \mathrm{rad} / \mathrm{s}$.

Key Concepts

Natural Frequency of a Mass-Spring System

This concept describes how a mass connected to a spring oscillates when disturbed. The natural frequency is determined by the mass and the effective spring constant, and is given by the formula ? = ?(k/m). Understanding this relationship is crucial for designing systems with specific oscillatory characteristics.

Effective Spring Constant

In systems with multiple springs, the overall stiffness is not simply the sum of the individual stiffness values unless the springs are arranged in particular ways. The effective spring constant results from the combination of individual spring constants, and has to be calculated based on whether the springs are configured in series (where 1/k_eff = 1/k1 + 1/k2 + …) or in parallel (where k_eff = k1 + k2 + …). This concept allows engineers to tailor the system’s stiffness to achieve desired vibrational properties.

Series and Parallel Combination of Springs

Springs can be arranged in series or parallel to modify the overall stiffness of a spring system. In a series configuration, the effective spring constant is lower than the smallest individual spring constant, while in a parallel configuration, it is the sum of the individual constants. Mastery of these concepts enables the design of complex spring systems that meet specific dynamic criteria, such as tuning the natural frequency of oscillatory systems.

Transcript

00:01 Okay, so springs are available in stiffness values of k equals 10, 100, and 1 ,000 newton per meter.

00:10 And we want to design a spring system that uses only these values so that a mass of 100 kilograms, when connected, has a frequency omega of about 1 .5 radiance per second.

00:26 Okay.

00:27 So the first thing that we're going to do is to figure out, our first step, is to figure out the equivalent stiffness.

00:40 In order to do that, i'm going to take my natural frequency and say that this is equal to the square root of k over m.

00:49 And the k here is the equivalent stiffness.

00:55 So m is 100 kilograms, omega is 1 .5.

00:58 So, k -e -q equals omega -n squared times m, which equals 1 .5 squared times 100, which equals 225 newton per meter.

01:18 So this is what we must have.

01:20 So this is a requirement.

01:25 This is what we must have.

01:31 Okay.

01:32 Now, we have springs that are available in stiffness values of 10 ,100 ,000.

01:40 So there are various ways of doing this.