Relationship between frequency and length of pendulum

Click For Summary
The discussion centers on the relationship between the frequency and length of a pendulum, highlighting the experimental equation f = 1.5L^{-0.5} derived from a log-log graph of frequency versus length. Participants clarify that this experimental equation differs from the theoretical equation f = 1/(2π√(L/g)), which is based on the pendulum's period. The significance of the experimental equation lies in its ability to validate the theoretical model through observed data. While both equations describe the same phenomenon, they serve different purposes in understanding pendulum dynamics. Ultimately, the theoretical equation can be used to assess the accuracy of the experimental findings.
Ace.
Messages
52
Reaction score
0

Homework Statement



We calculated times of the periods of varying pendulum lengths. (20cm, 40cm, 60cm, 80cm). Then the frequency was calculated for each length and then a frequency-length graph was made. Since the graph is an exponential relationship we graphed our values on a log-log chart. Then we found the equation y = kxn, where k is the value of y where x = 1, and n is the slope.

the following equation was found from the log-log chart, wehre f is the frequency and L is the length:
f = 1.5L^{-0.5}

What is the equation that relates frequency to length in a pendulum?



Homework Equations


T = 2\pi\sqrt{\frac{L}{g}}
f = 1.5L^{-0.5}


The Attempt at a Solution


I can come up with the equation for f from using the first equation

\frac{1}{f} = 2\pi\sqrt{\frac{L}{g}}
f = \frac{1}{2\pi\sqrt{\frac{L}{g}}}

But my issue is does the equation I found for my log-log chart (f = 1.5L^{-0.5}) play any role in finding the relationship? What is the significance of this equation? Would there be a way to derive f = \frac{1}{2\pi\sqrt{\frac{L}{g}}} using it?
 
Physics news on Phys.org
No, the equation you're finding is the theoretical one, and the equation you found from the data is the experimental one. They don't really have anything to do with one another, except that the theoretical one can verify the quality of the experimental one.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 3 ·
Replies
3
Views
708
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
1
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
37
Views
3K
Replies
28
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K