T^2 vs. L and T^2 vs M' Graphs

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In summary, the expected slope for the T^2 vs L graph is (4pi²/g) and for the T^2 vs M' graph, it is (4pi²/k). The calculated slope for T^2 vs L is 3.223 and for T^2 vs M' is 3.616. To calculate g, use the formula g = 4pi²/slope. The experiment involves a pendulum and an oscillating spring. M' represents the total mass while M is just the added weight. The equation for T^2 is (4pi^2m)/k, where m is the mass and k is a constant.
  • #1

Homework Statement


What is the expected slope of the line? What was the actual slope of the line of best fit? Calculate the gravity constant g from the slope of your graph

Homework Equations


k = (Mg) / (y_0 - y)

4pi^2/g x L = T^2

The Attempt at a Solution



I understand how to acquire gravity using the second equation for T^2 vs L. But I don't have a clue what my expected slope should be for either graph. My calculated slope for T^2 vs L is 3.223 and for T^2 vs. M' 3.616
 
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  • #2
If you graph y vs x and the formula is y = mx,
then the slope is expected to be m.
If you graph T² vs L and the formula is T² = (4pi²/g) x L
then the expected slope is (4pi²/g). The calc for g would then be
g = 4pi²/slope

I don't see a formula relating T² and M'. What is M'?
It looks like you might be doing a pendulum experiment?
 
  • #3
So that means my expected slope would be ~4? This was a pendulum experiment for T^2 vs L and and oscillating spring for T^2 vs M'. M' is the mass of our hook + spring + added weight, while M is just the mass of our weight added to the hook and spring.

Through my notes I found the equation T^2 = (4pi^2m)/k. So if I replace y=T^2 and x=m does that mean my slope is (4pi^2)/k?
 
  • #4
Yes, that is the idea. Good luck.
 
  • #5
.

As a scientist, the expected slope for both graphs should be approximately 4pi^2/g. This is because according to the equation T^2 = (4pi^2/g)L, the slope of the graph should be equal to 4pi^2/g. This is because the gravitational constant, g, is the proportionality constant between the period of oscillation (T) and the length of the pendulum (L). Therefore, the expected slope for both graphs should be close to this value.

To calculate the actual slope of the line of best fit, you can use the formula for slope: slope = (y2-y1)/(x2-x1). In this case, y represents T^2 and x represents L or M'. So the slope for T^2 vs L would be (T^2_2 - T^2_1)/(L_2 - L_1), and the slope for T^2 vs M' would be (T^2_2 - T^2_1)/(M'_2 - M'_1).

Once you have calculated the actual slope for each graph, you can use it to find the gravitational constant, g. Rearranging the equation for T^2 = (4pi^2/g)L, we get g = (4pi^2)/(slope). So for T^2 vs L, the calculated value of g would be (4pi^2)/(slope of T^2 vs L), and for T^2 vs M', it would be (4pi^2)/(slope of T^2 vs M').

Remember to use the units consistently throughout your calculations, as the units for g will depend on the units used for T^2, L, and M'. Once you have calculated the values for g, you can compare them to known values of the gravitational constant (9.8 m/s^2 or 32.2 ft/s^2) to see how close your results are. Any discrepancies could be due to experimental error or other factors that may have affected your data.
 

1. What is the purpose of a T^2 vs. L graph?

A T^2 vs. L graph is used to graphically represent the relationship between the period (T) of an oscillating object and its length (L). This graph can help determine the value of the gravitational constant, g, as well as verify the accuracy of theoretical equations.

2. How is a T^2 vs. L graph constructed?

To construct a T^2 vs. L graph, the length of the oscillating object is varied and the corresponding period is measured. The period is then squared and plotted on the y-axis, while the length is plotted on the x-axis. A best-fit line can be drawn through the points to determine the relationship between T^2 and L.

3. What does a straight line on a T^2 vs. L graph indicate?

A straight line on a T^2 vs. L graph indicates a linear relationship between T^2 and L. This means that the period is directly proportional to the square root of the length. This relationship is described by the equation T^2 = (4π^2/g)L, where g is the gravitational constant.

4. What is the significance of a T^2 vs. M' graph?

A T^2 vs. M' graph is used to graphically represent the relationship between the period (T) of an oscillating object and its reduced mass (M'). This graph can help determine the value of the gravitational constant, g, as well as verify the accuracy of theoretical equations.

5. How is a T^2 vs. M' graph different from a T^2 vs. L graph?

A T^2 vs. M' graph is similar to a T^2 vs. L graph, but instead of plotting the length on the x-axis, the reduced mass is plotted. The reduced mass takes into account the mass of the oscillating object and the mass of the object it is attached to. This graph is useful when dealing with systems of multiple masses.

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