Solving for linear mass density of a string

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To solve for the linear mass density of a string, the key equations involved are μ = M/L and the wave speed equation v = √(T/μ). The approach involves deriving expressions for frequency in terms of the number of segments and the slope of the frequency plot. By manipulating these equations, the final expression for linear mass density can be represented as μ = T/((df/dn) * 2L)², where df/dn is the slope of the frequency vs. number of segments plot. This method effectively incorporates the tension and length of the string into the calculation of linear mass density.
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HELP: Solving for equation of linear mass density of a string

Homework Statement


You will plot frequency vs. number of segments and determine a slope. Write the expression which will allow you to solve for the linear mass density of the string in terms of L, T, and the slope of your plot

Homework Equations


1) μ=M/L
2) T=2∏√(m/k) ?
3) λ=2L/n
4) df/dn=?
5) k=2∏/λ

The Attempt at a Solution


So what I've done is take equation 2 solve for m which is m=(T/2∏)2*k, with substitution from equation 5, then substitution from equation 3 to get the equation:
m=(T2*n)/(4∏L) which I substituted into equation 1 to get:
μ=(T2*n)/(4∏L2)
So so far I have the equation in terms of L and T but not in terms of the slope of my plot which would be df/dn.

Thanks!
 
Last edited:
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So I completely redid the problem a different way and I think this is right. Let me know if it is or not:
Equations:
1) v=√(T/μ)
2) v=(2Lf)/n

Attempt at solution:
Solve equation 1 for μ=T/v2, then solve equation 2 for f=vn/2L. Then derive f equation to get in terms of df/dn (slope of graph)= v/2L, solve for v, v=(df/dn)*2L and plug into the equation for μ to get μ=T/((df/dn)*2L)2.
This works right?
 
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?

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