Question

Use the data table below to answer the three parts of this question. Each part will be worth 10 points. mass length (m) Amp (degrees) time 10 swings (sec) T (sec) T (sec) 0.50 200 10 14.02 1.40 1.96 0.50 20 13.98 1.40 1.96 0.50 10 13.96 1.40 1.96 0.50 200 500 500 200 200 20 13.92 1.39 1.93 0.75 10 16.91 1.69 2.86 0.75 20 1.71 2.92 17.11 17.04 0.75 500 10 1.70 2.89 0.75 20 16.95 1.70 2.89 1.00 10 20.01 2.00 4.00 500 200 200 500 500 20 1.00 1.00 19.98 2.00 2.00 4.00 4.00 10 20.04 1.00 20 20.12 2.01 4.04 1.25 10 22.38 2.24 5.02 1.25 200 200 500 20 22.46 2.25 5.06 1.25 10 22.51 2.25 5.06 1.25 500 20 22.43 2.24 5.02 0.50 m 0.75 m Avg. T 1.40 $ 1.70 s 1.00 m 1.25 m 2.00 $ 2.25 s Avg. T2 1.95 s2.89 s? 4.01 s2 5.04 s 1. Use one data range statement to write a relationship statement between period and length. (use the average period and length information) 2. Use two data range statements to write a relationship statement between period and mass. (use same length and amplitude for your comparison) 3. Use two data range statements to write a relationship statement between period and amplitude. (use same length and mass for your comparison)

Answer 1

1.

We need a relationship between the period and the length.

Using the average period and lenth information, we tabulate the followring

length ( m )	T2 ( s2 )
0.50	1.95
0.75	2.89
1.00	4.01
1.25	5.04

Let us calucate the slope of length vs T² .

We use two sets of points from the above table,

Set 1:- ( length , T² ) = (0.50, 1.95) and (1.25, 5.04)

$Slope_1 = \frac{5.04 - 1.95}{1.25-0.50} = 4.12 \hspace{0.2cm} s^2/m$

Set 2:- ( length, T² ) = (1.25, 5.04) and (1.00, 4.01)

$Slope_2 = \frac{5.04 - 4.01}{1.25-1.00} = 4.12 \hspace{0.2cm} s^2/m$

Since the slope is same, we can say that length vs T² curve follows a straight line.

Thus the relationship between period and lenth is,

$T^2 \propto l$

Τα

So $period \propto \sqrt{length}$ , that is, period is proportional to square root of length.

2.

We need a relationship between the period and the mass (using same length and amplitude)

Using the following two data ranges,

Data range 1:- We choose length = 0.50 m and amplitude = 10 degrees.

So, the two sets are (period, mass) = (1.40, 200) and (1.40, 500).

We can see that even if the mass is changed, the period doesn't change.

Let us take another data range to confirm this statement.

Data range 2:- We choose length = 1.25 m and amplitude = 20 degrees.

So, the two sets are (period, mass) = (2.24, 200) and (2.25, 500).

We can see that even if the mass is changed, the change in period is negligible.

Thus we can conclude that the time period is independent of the mass.

3.  

We need a relationship between the period and the amplitude (using same length and mass)

Using the following two data ranges,

Data range 1:- We choose length = 0.75 m and mass = 200 g.

So, the two sets are (period, amplitude) = (1.71, 20) and (1.69, 10).

We can see that even if the amplitude is changed, the change in period is negligible.

Let us take another data range to confirm this statement.

Data range 2:- We choose length = 1.00 m and mass = 500 g.

So, the two sets are (period, amplitude) = (2.01, 20) and (2.00, 10).

We can see that even if the amplitude is changed, the change in period is negligible.

Thus we can conclude that the time period is independent of the amplitude.