How Does Graphing Time Squared Reveal Pendulum Dynamics?

AI Thread Summary
Graphing time squared against length and mass is crucial for analyzing pendulum dynamics. The relationship shows that as the length of the pendulum increases, the period of oscillation also increases, while mass has no effect on the swing time. By plotting time squared versus length, a straight line can be obtained, allowing for the calculation of the slope. This slope is directly related to the local gravitational constant, g, through the equation T_{p}^2 = (4π²/g)L. Thus, the experiment effectively demonstrates how to derive the gravitational constant from the pendulum's properties.
QuarkCharmer
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Homework Statement


We did a lab, where, in the first part we timed the period of a pendulum with varying lengths. In the second part, we timed the period of a pendulum with varying masses.

I got the results that I expected to get. However, I do not understand two of the lab questions.

A.) Length: Go to Graphical Analysis and Graph Time vs. Length. Explain the effects of length on the time of the swing.

B.) Go to Graphical Analysis and Graph Time vs. Mass. Explain the effects of mass on the time of the swing.


Homework Equations


T_{p}=2π\sqrt{\frac{L}{g}}

The Attempt at a Solution



Clearly, mass had no effect on the time of the swing, and as the length of the string increased, so did the period of oscillation.

What I don't understand is what I am supposed to graph exactly? For both of those parts, the professor instructed us to use a graph of time squared vs. length and time squared vs. mass. and then gave us a subtle hint to solve for g.

I don't understand what she is getting at with this cryptic clue. What is the significance of graphing the data with time squared? How am I supposed to find our local gravitational constant from this graph? (Assuming that is what was implied).

Thanks
 
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QuarkCharmer said:

Homework Statement


We did a lab, where, in the first part we timed the period of a pendulum with varying lengths. In the second part, we timed the period of a pendulum with varying masses.

I got the results that I expected to get. However, I do not understand two of the lab questions.

A.) Length: Go to Graphical Analysis and Graph Time vs. Length. Explain the effects of length on the time of the swing.

B.) Go to Graphical Analysis and Graph Time vs. Mass. Explain the effects of mass on the time of the swing.

Homework Equations


T_{p}=2π\sqrt{\frac{L}{g}}

The Attempt at a Solution



Clearly, mass had no effect on the time of the swing, and as the length of the string increased, so did the period of oscillation.

What I don't understand is what I am supposed to graph exactly? For both of those parts, the professor instructed us to use a graph of time squared vs. length and time squared vs. mass. and then gave us a subtle hint to solve for g.

I don't understand what she is getting at with this cryptic clue. What is the significance of graphing the data with time squared? How am I supposed to find our local gravitational constant from this graph? (Assuming that is what was implied).

Thanks

Graphing time squared is essential.

presumably when you graph time vs length you get a curve - but what curve?

If it curves up, it could be y= x2; y = x3; y = tanx

If it curves the other way, it could be the start of y= sinx ; y = √x ; y = 3√x

The only graph you can interpret with confidence is a straight line.

graphing the square of time against length might yield a straight line.
 
I understand that it straightens out the curve into a line, so you can then find a best fit, or calculate the slope of that line, or just read the data. What I don't understand is how this length v. time^2 graph can be used to find the local gravitational constant.

I found the slope of the graph to be about 4 point something or other, and I have been trying to figure out how I can use that for anything relating to the goal of this experiment.
 
Hi QuarkCharmer! :smile:

From your relevant equation it follows that:
T_{p}^2=\frac {4\pi^2} {g} L
This means you should find a straight line through the origin.
The slope is \frac {4\pi^2} {g}
So:
g = {{4\pi^2} \over \text{4 point something or other}} = 9.81

See how nicely it fits the actual acceleration of gravity! :wink:
 
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