# The Spring Model Continued – Baiting

In the past two weeks, I worked on Second Order Differential Equations with constant coefficients and learned more about the Spring-Damper Model. In this blog, I will provide a brief recap of the basic knowledge, and then provide further analysis of the same model. If you are interested in my last blog, please visit here.

Recap of Critical Definition from my Last Blog:

1. The standard form of homogeneous second order differential equations is:

• In which y is a function of tt is the independent variable, and A, B are arbitrary constants. While analyzing the spring-damper model, the independent variable t represents time.

2. The characteristic equation is:

• To solve a homogeneous second order differential equation with constant coefficient, set

• Solve for r, then use  to find y.

3. The spring-damper model is in the form of

• m is the mass of the block, b is the damping constant, and k is the spring constant.
• If you wonder how we get this equation, please check the link above and read my last blog.

Oscillation:

After reviewing the basic knowledge, we are now ready to analyze the damping of the spring-damper model! In this case, we will first transform

into

Remember that mass never equal to 0, so these two equations are equivalent.

And then the characteristic equation becomes

We first need to analyze how the roots (r) behave in the characteristics equation. We can use the same analysis for quadratic equation:

In this blog, we will look at “underdamping” in detail. If you are interested in other two types of damping, please first try to analyze them yourself then visit here.

First, I want to look at the implications of b^2<4mk in general. This may mean three things:

1. The damping constant b is small, which means the damper is weak.
2. The spring constant k is large, which means the spring is strong.
3. The mass is large, which means the block is heavy.

Imagine either one of the three cases, then you will find that as long as the mass(m) is not too big to break the spring, the block will oscillate and goes back to its equilibrium position. Let’s now try to prove this assumption mathematically!

First, since we have, we can set

So that

Plug it back into x and take the real part, we have

Notice that since the value in cos and sin are the same, we can use the sinusoidal identities and transform the equation into

In which

Following this equation, we would find:

1. When t (time) approaches infinite, (position) approaches 0. Which means it goes back to its equilibrium position. This agrees with our assumption.
2. The position of the block versus time should somehow follow the cosine function, which means it has a period and oscillates up and down. This agrees with our assumption.

If assign arbitrary values to b, m, and and plot the function, we would find something look like:

Graph from MIT Open Course Ware

And this is how underdamping looks like!! And conventionally, you can imagine the damper as gravity, and thus the motion would look like this:

In this blog, we looked at how a block oscillate with a spring and a damper when the damper is weak. We also analyzed and found the math agrees with our common sense in real life. In fact, many of our observations have an origin in physics or mathematics. As long as we use deliberately use math as a tool, we will be able to explain them!

Starting now, I will be still working on the online course and building my knowledge. But I will also start to refer more to other models not covered in the course material. Hope you will find them interesting and meaningful!

References

Mattuck, A., Miller, H., Orloff, J., & Lewis, J. (2011, Fall). 18.03SC Differential Equations. Retrieved November 5, 2018, from MIT Open Course Ware website: https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/index.htm

Zhu, B. (2018, October 21). Spring Model – Baiting [Blog post]. Retrieved from Independent Seminar Blog: https://independentseminarblog.com/2018/10/21/spring-model-baiting/

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