How can I use log form to find the constant k in a given equation?

  • Thread starter Thread starter phzxc
  • Start date Start date
AI Thread Summary
To find the constant k in the equation T = kL^(3/2), the equation can be transformed into log form by taking logarithms of both sides, resulting in log T = log k + (3/2) log L. This transformation linearizes the relationship between T and L, allowing for a straightforward graphing approach. By plotting log T against log L, the slope of the line will be 3/2, and the y-intercept will represent log k. The value of k can then be determined by calculating the antilogarithm of the y-intercept.
phzxc
Messages
2
Reaction score
0

Homework Statement


So, I'm given the equation T = kL^3/2

Data: L= .9, .8, .7, .6, .5 and T=.558, .47, .375, .323, .26 (.9 goes with .558, etc)

I need to find the constant k by changing T=kL^3/2 into log form

Homework Equations





The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
"Rewriting the equation in log form" essentially means to take logarithms on both sides of the equation. So, we obtain
log T = log (k\,L^{\frac{3}{2}})​

I'm sure you can go on in further simplifying the expression?
 
Well, I see you've edited your post with the data.
To solve for k, what we are doing here is actually linearising the relation between T and L so that we can plot a nice straight line.
Simplifying the expression further, we get:
log\,T = log\,k + \frac{3}{2}log\,L​

Clearly, plotting log T against log L (values obtained from your data) will yield a gradient of 3/2 and a y-intercept of log k. This enables you to obtain the value of k.
 
how does making the relation linear allow me to find k?
 
phzxc said:
how does making the relation linear allow me to find k?
It allows you to plot a simple straight line graph in the form y = mx + c from which you can extract information from.
As I mentioned in my earlier post, plotting y (log T) against x (log L) will yield a gradient m (3/2) and a y-intercept c (log k). Obtain the y-intercept value from the graph, which is equal to log k, and solve from k from there.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top