The note that is three octaves above middle C is supposed to have a fundamental
ID: 1500326 • Letter: T
Question
The note that is three octaves above middle C is supposed to have a fundamental frequency of 2093 Hz. On a certain piano the steel wire that produces this note has a cross-sectional area of 7.85 10-7 m2. The wire is stretched between two pegs. When the piano is tuned properly to produce the correct frequency at 25.0°C, the wire is under a tension of 818.0 N. Suppose the temperature drops to -19.4°C. In addition, as an approximation, assume that the wire is kept from contracting as the temperature drops. Consequently, the tension in the wire changes. What beat frequency is produced when this piano and another instrument (properly tuned) sound the note simultaneously?
______ Hz
Explanation / Answer
f = f0 * sqrt [F0 + alpha(delta T) Y A / F0]
= 2093 * sqrt [818 + 12 * 10-6(5.6) 2 * 1011 * 7.85 * 10-7 / 818]
= 2106 Hz
Beat frequency = 2106 - 2093 = 13 hz
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