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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 densitymu and length L, determine the ratio opf tensions in the two strings . |
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Answer» Solution :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. |
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