## A Springy Thingy, Part 2

So last time, I presented the equation that describes the position of a mass on a spring:

The equation is

${x}\hspace{0.33em}{=}\hspace{0.33em}{A}\hspace{0.33em}\sin\left({\frac{180t}{\mathit{\pi}}\sqrt{\frac{k}{m}}}\right)$

where x is the position of the mass at a given time t in seconds, k is the spring constant, and m is the mass in kg. This was developed from the equation that describes the forces on the spring (gravity and the spring), and through calculus, out pops the equation above. This equation is the sine of stuff in brackets multiplied by a number A.

Even though the stuff in the brackets looks rather ominous, we are still just taking the sine of it and the sine only goes from -1 to 1. So the maximum extent of the mass is from –A to A. Now let’s look at the stuff in the brackets.

The 180 and 𝜋 are just there to change the rest of the expression so that you can press the sine button on your calculator in the “degrees” mode. Normally, when engineers model something like this, they use radians and not degrees. I have not explained what radians are yet so I’ve included an adjustment (the 180 and the 𝜋) so that you can continue to use degrees. I think I’ll explain radians in my next post.

The rest of the numbers, t, k, and m are the real meat of the model. For simplicity, I have started time at 0 seconds when the mass is at its rest position and is moving upwards in the postive direction. So you would expect the position of the mass to change with time and that is what the t in the expression does. The k and the m determine how fast or how slowly the mass oscillates. Let’s actually use some numbers instead of letters here for a specific mass and spring.

Now let’s assume the spring has a spring constant of 1 kg/s² (I’ll discuss these units in my next post), and the mass connected to it is 1 kg. That means the stuff in the square root sign (called a radical) is just 1 and the square root of 1 is 1. And let’s further assume that I start the spring moving by stretching the spring 5 cm from its rest position. So now, the position equation above simplifies to

${x}\hspace{0.33em}{=}\hspace{0.33em}{5}\sin\left({\frac{180t}{\mathit{\pi}}}\right)$

Starting at time as 0, you can choose various values of t, compute the stuff in the brackets, use the “SIN” button on your calculator which is in the “degrees” mode, and then multiply by 5. So for example, at t = 1 sec, 180/𝜋 is 57.2958. Taking the sine of that gives 0.8415 and then multiplying that by 5 gives 4.2073 cm. So at 1 sec, the mass is 4.2073 cm above its resting position. You can plot this point and many others to graph this, or you can be lazy like me and use a graphing calculator. The graph of the position of the mass versus time for this scenario is

So no surprise, a sine wave. Now remember when I first talked about sine waves, I talked about the wavelength. Here I have indicated the wavelength as 6.28 sec. When dealing with time, the wavelength is called the period and usually represented with the symbol T. The period is the length of time it takes for one full cycle of motion. So it takes the mass 6.28 seconds to make one complete bounce. It turns out that you do not have to graph the curve to find this:

${T}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\mathit{\pi}}{\sqrt{\frac{k}{m}}}$

Since our k and m are each 1, T in this case is just 2𝜋. Funny how 𝜋 keeps cropping up. Again, I’ll explain that in my next post on radians.

Associated with the period is something called frequency. The period is how long it takes for one complete cycle to occur, whereas the frequency is how many complete cycles occur in 1 second. Frequency is the reciprocal of the period and vice versa. That is f = 1/T. So for our mass, the frequency is 1/6.28 or 0.16 cycles per second. The term “cycles per second” is given a special unit called hertz which is abbreviated as hz. You may have heard this term before.

As you change the values of k and m, the values of T and f will change as well. If the spring gets stiffer (a higher k), you would expect the frequency to increase, that is it will bounce faster. You would expect a heavier mass to slow down the frequency and it does. I will leave it as an exercise for the student to check this using a graphing calculator or Excel.

A good simulation on the web that shows the effect of changing mass an spring constant is at https://www.physicsclassroom.com/Physics-Interactives/Waves-and-Sound/Mass-on-a-Spring/Mass-on-a-Spring-Interactive. This sets up the graph a bit differently than I do here, but the frequency changes are easy to see. Also, you can add damping to this which I did not include in this post to keep it simple, but you can play with that as well on this site.

## A Springy Thingy, Part 1

From my last three posts, I think we are ready to model the motion of a mass on a spring:

So, just like I did with the tennis ball, I will show you equations that describe this motion.

Now again, to develop these equations requires calculus, so I will just provide the final result. But before I do, just how does one begin modelling a physical process like this?

What is usually done, is to write down known equations that describe the forces acting on the mass. If you think about it, there are two: a force due to gravity and a force from the spring. There are other forces as well like resistance from the air, but as before, we will assume these to be zero to simplify the development.

Let’s first set up the picture. We have a weight on a spring. The weight has mass m. Now weight is different than mass, but on earth, the units are the same. So on earth, a 1 kg weight has a mass of 1 kg. But on the moon, the mass is still 1 kg but its weight is 0.165 kg because gravity is weaker there. We have a spring with a spring constant of k which is a measure of how stiff the spring is. The higher the value of k, the stiffer the spring.

To start the mass moving, we have stretched it A centimeters down from its resting position, then let go. We set up a one dimensional coordinate system where the rest position of the mass is 0 and up is positive.

The force due to gravity is –mg where m is the mass and g is the acceleration due to gravity. From my post on the tennis ball, remember that g is 9.8 m/s². It’s negative because the force is acting in the down direction. This comes from Isaac Newton’s second law that says that force is equal to mass times acceleration. That is, F = ma. The force due to the spring comes from something called Hooke’s Law: F = kx where k is the spring constant and x is the amount that the spring is stretched (negative) or compressed (positive) from the resting position.

So the force equation for this setup is:

F = ma = kxmg

This is the equation engineers start with before they do calculus on it. So now in this post, this is the part where a miracle happens, and I’ll give you the final result.

So the equation that shows where the mass is at a certain time is below where t is time in seconds. It is assumed that time starts (that is t = 0) when the mass is travelling upwards and is at the 0 position:

$%Translator MathMagic Personal Edition Mac v9.41, LaTeX converter, 2019.1.28 09:06 {x}\hspace{0.33em}{=}\hspace{0.33em}{A}\hspace{0.33em}\sin\left({\frac{180t}{\mathit{\pi}}\sqrt{\frac{k}{m}}}\right)$

In my next post, I’ll dissect this a bit and put some actual numbers in it and plot the results.

## What Goes Up …

I was a mediocre student in maths before I started at a university. But in university, I discovered the amazing things that maths can do. Maths can be used to describe all sorts of physical processes and this can be used to control those processes. Things like the cruise control in your car or the flight controls of an aircraft or spacecraft need a mathematical model of the thing being controlled. One of the first examples used to introduce students to modelling (that is mathematically describing something), is the simple act of throwing a ball into the air.

Now to start modelling this process from scratch requires calculus which I haven’t covered yet. But I will give you the final result and we will work with that.

If a tennis ball is hit straight up in the air, there are two main forces acting on it after it leaves the racquet: gravity and drag from the air. Though the air drag can be modelled as well, it complicates the model so it is usually assumed to be negligible when introducing this to students. The effect of gravity is to reduce the initial speed given to the ball by 9.8 m/s every second. So if the ball has an initial speed of 60 m/s, after 1 second its speed is 60 – 9.8 = 50.2 m/s. After the next second, its speed is 50.2 – 9.8 = 40.4 m/s, and so on. By the way, I am using metric units here. The same thing can be done with American units where the effect of gravity reduces the speed of the ball by 32 ft/s every second. But as most of the world uses metric, we’ll stay with that.

So let’s stick with the initial velocity of the ball as 60 m/s when it leaves the racket. Now the first thing to do is agree on a coordinate system. It’s natural to agree that up is positive and down is negative. We’ll also agree that time starts at 0 as soon as the ball leaves the racket and that the height in meters at the point where the ball leaves the racquet is also 0. We’ll call this the ground level. So using calculus with the force of gravity and the initial speed of the ball (ignoring air drag), we can get three equations that describe the motion of the ball: one equation for its acceleration, one for its velocity, and one for its height from the ground. With a = acceleration, v = velocity, h = height, and  t = time in seconds, these equations are:

$\begin{array}{l} {{a}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}}\\ {{v}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}}\\ {{h}\hspace{0.33em}{=}\hspace{0.33em}{-}{4}{.}{9}{t}^{2}\hspace{0.33em}{+}\hspace{0.33em}{60}{t}} \end{array}$

With these equations, we can answer the following questions:

1. When will the ball hit the ground?
2. How fast is the ball travelling when it is at its maximum height?
3. How high does the ball go?
4. How fast is the ball going when it hits the ground?

Before I answer these questions, let’s look at the graph of the height equation. Now before, we talked about x and y values on a graph. But for physical processes, we can use different letters that are more meaningful. Instead of x, we will use t for time, and instead of y, we will use h for height. The graph of this equation is below:

The curve is an upside down parabola. The graph goes on forever below the t-axis, but I’m only show the part of the graph that makes physical sense.

The graph and the equation make sense at t = 0 seconds as h is 0 on the graph at t = 0, and if you let t = 0 in the equation, you also get h = 0.

So you can see that the ball goes up, reaches a maximum height, then falls back to the ground. To answer question 1, from the graph, it looks like the ball hits the ground a little over 12 seconds because that is where the graph shows h = 0 again. To find the exact value, we must set the height equation equal to zero, and find the times when h = 0. This will require us to factor the equation and use the null factor law (explained in a previous post):

$\begin{array}{l} {{h}\hspace{0.33em}{=}\hspace{0.33em}{-}{4}{.}{9}{t}^{2}\hspace{0.33em}{+}\hspace{0.33em}{60}{t}\hspace{0.33em}{=}\hspace{0.33em}{t}{(}{-}{4}{.}{9}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}}\\ {\Longrightarrow\hspace{0.33em}{-}{4}{.}{9}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{-}{4}{.}{9}{t}\hspace{0.33em}{=}\hspace{0.33em}{-}{60}}\\ {\Longrightarrow{t}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{60}}{{-}{4}{.}{9}}\hspace{0.33em}{=}\hspace{0.33em}{12}{.}{245}\hspace{0.33em}{\mathrm{seconds}}} \end{array}$

So the ball travels for 12.245 seconds before it hits the ground (that’s a powerful tennis player!). Notice that there is also another solution, t = 0, which is when the ball is initially hit.

Question 2 can be answered by the physics of the problem and this answer will help us answer question 3. As explained, the ball has an initial velocity of 60 m/s, but is continually slowing down due to gravity. When it reaches its maximum height, the ball reverses direction then goes back down. At the maximum height, the velocity is 0 because velocity is positive going up and negative going down, so it must be 0 right when the ball reverses direction.

Question 3 can be answered using the answer to question 2. If we set the velocity equation to zero, that will give us the time that the ball is at maximum height. We then use this time in the height equation to find the height at that time:

$\begin{array}{l} {{v}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{-}{9}{.}{8}{t}\hspace{0.33em}{=}\hspace{0.33em}{-}{60}}\\ {\Longrightarrow\hspace{0.33em}{t}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{60}}{{-}{9}{.}{8}}\hspace{0.33em}{=}\hspace{0.33em}{6}{.}{122}\hspace{0.33em}{\mathrm{seconds}}}\\ {{h}\hspace{0.33em}{=}\hspace{0.33em}{-}{4}{.}{9}{(}{6}{.}{122}{)}^{2}\hspace{0.33em}{+}\hspace{0.33em}{60}{(}{6}{.}{122}{)}\hspace{0.33em}{=}\hspace{0.33em}{183}{.}{67}\hspace{0.33em}{\mathrm{meters}}} \end{array}$

We also could have solved this by noticing that a parabola is symmetric and the maximum height would occur halfway between 0 and 12.245 seconds. This also would have given us t = 6.122 seconds to use in the height equation.

The last question is answered by using the time that the ball hits the ground in the velocity equation:

${v}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}{(}{12}{.}{245}{)}\hspace{0.33em}{+}\hspace{0.33em}{60}\hspace{0.33em}{=}\hspace{0.33em}{-}{60}\hspace{0.33em}{\mathrm{m}}{/}{\mathrm{s}}$

Notice that the velocity is the same as when the ball was first hit except it is negative since it is now going down instead of up.

Isn’t it amazing how we can find out all sorts of things about throwing a ball without actually doing it!