Standing Wave Problem, What is the mass of the string?

AI Thread Summary
The discussion revolves around calculating the mass of a string and the required tension for specific vibrational modes. For part a, the mass of the string was calculated as 0.0001 kg, with the approach deemed correct despite a lack of unit specification. In part b, the calculation for the tension needed for the string to vibrate in three segments at 150 Hz was based on the correct formula, but it was noted that the mass per unit length used should match the earlier calculation. The overall method appears valid, but clarity on unit consistency and values used is essential. Double-checking calculations and consulting with peers for confirmation is recommended for accuracy.
timtng
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When the tension is 18 N, a string 2.00 m long has a fundamental frequency of 150 Hz.
a.) What is the mass of the string?
b.) With what tension must the string be stretched so that it vibrates in three segments at 150 Hz?

This is what I came up with:
a.) f=v/2L, v=2Lf=2*2*150=600 m/s
v=sqrt(T/μ), v^2=T/μ, μ=T/v^2=18/600^2=.00005 kg/m
M=2*.00005= .0001 kg

b.) L=3λ/2, 2=3λ/2, λ=4/3
v=fλ=150*4/3=200 m/s
v^2=T/μ, T=μ*v^2=.00005*200^2= 2N

My friend kept on saying that I did this problem incorrectly.
Is my friend right?
 
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Your basic approach looks correct. (Didn't check the numbers.)
 
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It is difficult to determine if your solution is incorrect without knowing the specific instructions or context of the problem. However, your calculation for the mass of the string (part a) appears to be correct based on the given information. The only potential issue is that you did not specify the units for the mass, which should be in kilograms (kg).

For part b, it seems like you are using the correct formula, T = μv^2, to determine the tension needed for the string to vibrate in three segments at 150 Hz. However, you did not mention what value you used for the mass of the string (μ) in this calculation. It should be the same value that you calculated in part a, which is 0.00005 kg/m. If you use this value, then your solution for part b is correct.

It is always a good idea to double check your calculations and make sure you are using the correct units for each quantity. If you are still unsure, it may be helpful to consult with your teacher or classmate to clarify any confusion.
 
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