In the last section we introduced various parts of the language of science, units, graphs, and formulas and indicated that there are four steps to understanding formulas. In this section we will discuss that last two steps, and illustrate them with the same example of gas in a container, where:

p = nRT/V.

Again, consider the pressure of a gas in a container:

p = NRT/V,

where N is the number of moles of gas, R is the gas constant, T is temperature and V is the volume of the container. Notice that you might not understand some of these terms, like the number of moles of gas or the gas constant. However, the great thing about dependencies is that you do not need to know or understand everything to answer some useful questions. This is really an important goal of physics, to make interesting observations or predictions even in the face of incomplete understanding.

A typical problem starts with knowing pressure in the container for particular values of N, T, and V. If we change on of these quantitities, we want to know how the pressure will change. For example, the pressure is 35 pounds/square inch (PSI) (about the air pressure in the tires on a car). What is the new pressure if I increase the temerature by a factor of 1.25?

Here is how to proceed: the original pressure is p_original = 35 PSI.

So, p_original = NRT/V

The new pressure will be: p_new = NR(1.25T)/V because we have replaced T with 1.25T. In other words, we have increased the temperature by a factor of 1.25.

The next step is to isolate the factor of 1.25 that you introduced into the formula:

p_new = NR(1.25T)/V = (NRT/V)(1.25)

What is left in the bracket is just the original frequency, p_original, so we can substitute that into the expression for the new pressure:

p_new = (p_original)(1.25)

So, the new pressure is (35 PSI)(1.25) = 43.75 PSI.

We will now consider an example that is much more important for this class: the frequency of a vibrating string.

Step 1: What does the frequency of a string depend on?

The length of string, L, the tension or how tightly it is pulled, T, and the density or thickness of the string U.

Step 2: What is the formula for the frequency of the string? At this point, we don't know, but let's consider the following:

, , , , ,

Step 3: Qualitative dependence.

Using what you know about the pitch of a note on a string instrument, which of the formulas above can work?

Step 4: Quantitative dependence.

How does frequency depend on length?

If the frequency is 500 Hz and I triple the length of the string, what is the new frequency?

, so .

Or = 167 Hz.

Questions on quantitative dependences can also be asked in the reverse way, and this is sometimes more important. For example, we can ask, by what factor do I have to change the length of a string to make its frequency increase by a factor of 2? The simplest way to solve this is to make a guess as to what change in the length will work and try it out. If it does not give the correct result, you can usually then predict what will work. The nice thing is that you can always check your answer to make sure it works.

The last aspect of formulas, briefly mentioned above, is verifying them experimentally. This is important when you have a formula, but you don’t know from where it came. Rather than just accepting it as correct, you can test it in the lab.

For example, we want to test the formula for the frequency of the string. It is hard to test the whole formula at once, so we will look at one part of it: the dependence of frequency on the length of the string. Interestingly, we can do this without knowing anything about the other quantities in the formula. We start with:

.

Since we are interested in frequency and length, we can move them to the same side of the equation:

.

The important thing to notice is that the product of f× L is constant. The value of the constant does not concern us; we just note that it is a constant. This gives us a way to test the dependence. We set up a vibrating string in the lab and measure the frequency for different values of the length of the string (while keeping everything else the same!):

From this example data, we can calculate various combinations of f and L. If we find a combination that is a constant, even as we change the length of the string, we know that we have found out how the frequency depends on the length of the string. In this case, it will turn out that f× L = constant, and so, f = constant/L, as it should. However, we did not have to rely on the formula to reach this conclusion. We could have learned this simply from the data.

This experiment does not tell us everything about the frequency of a string, but we do learn something. We can also vary the tension in the string and keep the length and thickness constant to study how the frequency depends on the tension of the string.

The most significant conclusion from this section is that we do not need to know or understand every aspect of a formula to be able to get some useful information from it. We can answer certain questions without knowing everything. If it was not for this, physics and science would never have progressed as no one person understands everything!