23-Feb-2015: Obtaining relationship between mass and period for an inertial balance

Purpose:

This lab is to finding equation that relates period (T) to mass (m) for inertial balance. We will be using several different identified masses and a few unknown masses to obtain the the unknown constant "A" and "n" in the equation T= A(m+Mtray)^n.

Procedure:

1. By using C-clamp, attach the inertial balance to the end of the table and another separate clamp stand holding the Vernier photogate sensor. At the end of the balance, attach a piece of masking tape. As when the balance oscillates, the tape doesn't touch the Photogate.
2. Mass is added continuously on top of the inertial balance, each time 100g, starting from zero to 800g. Thereby, 9 trials of period are collected into Logger Pro. The value of period that should be considered part of the data is the last period value when "collection" stops.
3. In addition, we also recorded the period of two unidentified mass from stapler and cellphone. Gravitational masses of these two objects were obtained from a scale and will be used to compare to their inertial masses later on.

m=800g are placed on the inertial balance, while the photogate collects data

Stapler of gravitational mass= 369g

Data and Explanation:

_ In order to make our assumed equation, T= A(m+Mtray)^n, looks more like a liner equation, we have to take natural logarithm of both side:

lnT= lnA + n*ln(m+Mtray);
By making extra columns in Logger Pro for lnT, lnA, (m+Mtray), and ln(m+Mtray), using Parameter command, we can make a new graph having y-axis= lnT and x-axis= ln(m+Mtray).

As promised, we will get a relatively linear line. And by using "Linear fit" command, it will also give a correlation of the graph. The best correlation is 1.

This is when we can start playing with the value of Mtray. In Data/User Parameter menu, make a guess on value of Mtray and see if the correlation gets better. Having three decimal places is recommended as the value we obtain is more of range rather than an exact value of Mtray.

Here are the two screenshots of low and high-value of Mtray,

Assumed low value of Mtray= 0.309kg, Correlation=0.9997

Assumed high value of Mtray= 0.358kg, Correlation=0.9997

 From these two linear fits we have:
Low value: Mtray= 0.309kg; n=m= 0.6936; b=lnA= -0.4361
=> A = e^(-0.4361)= 0.6466

Hence, T = 0.6466*(m+0.309)^0.6936

High value: Mtray= 0.358kg; n=m=0.7530; b=lnA= -0.4696
=> A= e^(-0.4696)= 0.6253

Hence, T = 0.6253*(m+0.358)^0.7530

_ Identifying inertial masses of phone and stapler:

Phone's gravitational mass= 0.161kg; T= 0.383438 s
Stapler's gravitational mass= 0.369kg; T= 0.495054 s

The following is the calculations to finding inertial masses of phone and stapler:

calculations for inertial masses and table of comparison

From the table, it seems that the gravitational masses of both object are close but not in the range of their inertial masses.

After obtaining the values for "A" and "n", we use the equation T= A(m+Mtray)^n  again to graph with the help from Power Law fit command. The x-axis represents (m+Mtray), while y-axis is for T. Since Power Law fit is among various commands in Curve fit, a curve is expected. However, the graph below appears to be a good linear fit.

From the above graph:

Mtray= 0.358kg; Power Law Fit for T and (m+Mtray)

Mtray = 0.309kg; Power Law fit

From the above graphs, A and n are relatively close to the ones obtained from log-log fit.

Conclusion:

This lab not only finds the relationship between period and mass, it also gives an insight that mass can be measured using inertial balance(given Mtray); even though the process takes longer than just using the scale. This lab also teaches the use of applying logarithm in order to get a linear equation from exponential equation. Last but not least, the Power Law Fit allows for a shortcut solution to finding constant "A" and power "n" (by just guessing value of Mtray until the graph seems linear).