Relating y=mx+c and T^2=kd^3+4pi^2l/g

  • Thread starter Thread starter Natalie Morris
  • Start date Start date
Click For Summary
The discussion centers on the relationship between the linear equation y=mx+c and the equation T^2=kd^3+4pi^2l/g, with participants questioning their connection. The first equation represents a linear relationship in two-dimensional space, while the second relates to the period of a pendulum and involves constants k and g. The context of the second equation arises from a pendulum lab experiment aimed at determining values for k and g. Participants express confusion over the relevance of the two equations and seek clarification on their comparison. Ultimately, the discussion highlights the need for context to understand how these equations might be related.
Natalie Morris
Messages
2
Reaction score
0

Homework Statement


Can someone please tell me how these two equations are related

Homework Equations


y=mx+c; T^2=kd^3+4pi^2l/g[/B]
 
Physics news on Phys.org
The first is just the generic equation for a line in 2-D space. The second does not seem to be related to it at all. Are we supposed to just guess where you got these and why you want to compare them? That is, should we just guess what the context of your question is?
 
Natalie Morris said:

Homework Statement


Can someone please tell me how these two equations are related

Homework Equations


y=mx+c; T^2=kd^3+4pi^2l/g[/B]
Hi NM. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

I don't recognize the kd^3 term. Where did this equation for T^2 come from?

This old thread may hold some answers: https://www.physicsforums.com/threads/shm-pendulum-length-gravity-question.529318/
 
Last edited by a moderator:
Well we got a pendulum lab to perform and had to relate the period T^2 to the distance of a yielding support d^3. That equation was provided to us to use to determine the values of k and g in the equation, where k was a constant and g was acceleration of free fall
 
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 15 ·
Replies
15
Views
4K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 10 ·
Replies
10
Views
8K
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 1 ·
Replies
1
Views
10K