Middle `C` on a piano has a fundamental of `262 Hz`, and the first `A` above middle `C` has a fundamental frequency of `440 Hz`.
a. Calculate the frequencies of the next two harmonics of the `C` string .
b. If `A and C`, strings have the same linear mass density ` mu` and length `L`, determine the ratio opf tensions in the two strings .

Explanation

Remember that the harmonics of a vibrating string have frequencies that are related by integer multiples of the fundamental .
This first part of the example is a sample substitution problem.
Knowing that the frequency is ` f_(1) = 262 Hz` , find the frequencies of the next harmonics by multiplying by integers :
` f_(2) = 2f_(1) = 524 Hz`
`f_(3) = 3f_(1) = 786 Hz`
b. This part of the example is more of an analysis problem than is part (a).
Use Eq. (iii) to write expression for the fundamental frequencies of the two string.
` f_(1 A) = (1)/( 2 L) sqrt((T_(A))/( mu)) and f_(1) C = (1)/(2L) sqrt (( T_(c))/(mu))`
Divide the first equation by the second and solve for the ratio of tensions
`( f_(1)A)/( f_(1) C) = sqrt((T_(A))/(T_(C)))`
`(T_(A))/( T_(C )) = ((f_(1) A)/( f_(1) C))^(2) = (( 440)/( 262))^(2) = 2.82`
If the frequencies of piano strings were determined solely by tension . This result suggests that the ratio of tensions from the lowest string to the highest string on the piano would be enormous . Such large tensions would make it difficult to design a frame to support the strings . In reality , the frequencies of piano strings vary due to additional parameters , including the mass per unit length and the length of the string.