Find mass per unit length of a string graphically

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The discussion focuses on determining the mass per unit length (μ) of a string graphically, using the equation μ = T/v^2. The user has calculated μ as 0.00089 kg/m based on their experimental data, but they are experiencing a significant discrepancy of about 23% compared to the expected value. They have graphed velocity against the square root of tension, yielding a slope of 33.5, which seems inconsistent with their calculations. Additionally, there is confusion regarding the calculation of the string's weight and length, as the user notes that the ratio of weight to length does not match their μ value. The thread highlights the importance of accurately determining experimental parameters and graphing techniques in physics experiments.
jbumbes
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Homework Statement


frequency, Tension, mass, mode of vibration

Homework Equations


v = sqrt(T/μ)

The Attempt at a Solution


length of string is 2.14m
weight .00247kg
mass per unit length (μ) .00089

However, I need to confirm this graphically. I solved for mass per unit length
μ = T/V^2
or sqrt(μ) = sqrt(T)/v
however, I have already graphed this and the slope is 33.5. With velocity on the y-axis and sqrt(T) on the x which seems way too off. Basically I have to find the experimental μ graphically.
 
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That gives an error of about 23% then - the difference between the experimental and real value for linear density?
You did take the wavelength to be the twice the distance between two adjacent nodes?
Is the string running straight from the top of the pulley to the other end where it is fixed or tied to the vibrator?
 
jbumbes said:
length of string is 2.14m
weight .00247kg
mass per unit length (μ) .00089
This in itself is strange. Can you explain how you calculated this ?
 
That is the experimentally determined value, which then gives a 23% difference between the two.
 
I mean that 2.47/2.14 is not 0.89 but 1.15 gram/meter
 
The 0.00089 value is from his gradient
 
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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