Linear Mass Density: Find 8.9866e-4 m/s to g/m

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Homework Help Overview

The discussion revolves around finding the linear mass density using a relationship derived from the equation for velocity in terms of tension and linear density. The original poster attempts to determine the units of their calculated linear mass density and seeks clarification on converting these units.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the units of velocity and tension, questioning how taking the natural logarithm affects the units. There is a focus on ensuring consistency in units across the equations used.

Discussion Status

Some participants have provided insights into the units of tension and velocity, suggesting that the original poster clarify the units of linear mass density. The conversation is ongoing, with multiple interpretations of the unit relationships being explored.

Contextual Notes

There is a mention of experimental data and specific values obtained, but the discussion is constrained by the need for clarity on unit conversions and the implications of logarithmic transformations on those units.

brianna0404
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Linear Mass Density!

Homework Statement



Find linear mass density using a graph ln(v) vs ln(T), from Velocity=squareroot Tension/linear density, take natural log ln(v) = 1/2 ln(T)-1/2ln(m). our professor told us to find it this way cause 1/2ln(m) is the y-intercept of the best fit line of five trials

Homework Equations


so y-intercept=-1/2ln(m). for y intercept I got 3.5073 from an experiment done in lab.

The Attempt at a Solution


I got 8.9866 * 10^-4 for linear mass density. but after all that my main question is what are the units? would they be m/s? if so how do I convert it to g/m?

Thanks!
 
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it depends on units in which v and T are given. you have to extract units from formula Velocity=squareroot Tension/linear density.

Units of right and left parts of equation must be the same.
 


the units for tension are Newtons and the units for velocity are hertz times meters, which is just m/s. So i didn't know if taking the natural log of tension and velocity changes the units or are they the same?
 


logarithm equation and equation without logarithm are eqivalent
if [v]=m/s, it doesn't mean that [ln(v)]=m/s of course. They are two different numbers representing the same velocity and can't have the same dimension.
but if you are consistent with formulas you have in your calculations, you will get right units from the initial equation.

because

ln(v) = 1/2 ln(T)-1/2ln(m)

is the same as

ln(v)=ln(sqrt(T/m))

is the same as

v=sqrt(T/m), m=T/v*v.

So, if [T]=N, [v]=m/s, you can easily extract [m]. And it is not m/s.
 

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