Asked by Anonymous
A piano tuner stretches a steel piano wire with a tension of 765 N. The steel wire has a length of 0.400 m and a mass of 3.00 g. What is the frequency f_1 of the string's fundamental mode of vibration? What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 9781 Hz?
Answers
Answered by
bobpursley
http://hyperphysics.phy-astr.gsu.edu/Hbase/waves/string.html
Answered by
Billy
In Part A of this particular problem you are solving for the Fundamental Frequency of the string. Use the equation:
F(fundamental)= [1/(2 * L)] * sqrt[(tension force/mu)]
L= length of string= .4 meters
mu= mass/length=.003 kg/.4=.0075 kg/m
tension force=765 N
plug in knowns and F= 399.2 Hz
*************************************
For part B you must solve for the integer multiple, which the problem calls "n". The formula is as follows:
F1= n * F(Fundamental Frequency)
F1=9781 Hz
solve for "n" and n= 24.5....round down so that n=24.
F(fundamental)= [1/(2 * L)] * sqrt[(tension force/mu)]
L= length of string= .4 meters
mu= mass/length=.003 kg/.4=.0075 kg/m
tension force=765 N
plug in knowns and F= 399.2 Hz
*************************************
For part B you must solve for the integer multiple, which the problem calls "n". The formula is as follows:
F1= n * F(Fundamental Frequency)
F1=9781 Hz
solve for "n" and n= 24.5....round down so that n=24.
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