The B-string of a guitar is made of steel (density 7800 kg/m 3 ), is 63.5 cm long, and has diameter 0.406 mm. The fundamental frequency is f = 247.0 Hz.
f if the temperature of the string rises to 29.5ºC. The steel string has a Young's modulus of 2.00 × 10 11 Pa and a coefficient of linear expansion of 1.20 × 10 –5 (Cº) –1 . Assume that the temperature of the body of the guitar remains constant. Will the vibration frequency rise or fall? [Out of Tune]
F, the frequency f changes by a small amount 
f. Show that 
= 
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
For a string, f n = 
and in this case, n = 1. Rearranging this and solving for F gives
F = µ4L 2 f 2 . Note that µ = 𝜋 r 2 ρ,
so µ = 𝜋 (0.203 × 10 –3 m) 2 (7800 kg/m 3 ) = 1.01 × 10 –3 kg/m. Substituting values, d
F = (1.01 × 10 –3 kg/m) 4(0.635 m) 2 (247.0 Hz) 2 = 99.4 N.
To find the fractional change in the frequency we must take the ration of 
f to f:
Now divide both sides by the original equation for f and cancel terms:
From Section 17.4, 
F = – YαA 
T, so
F = –(2.00 × 10 11 Pa)(1.20 × 10 –5 /Cº) × ( 𝜋 (0.203 × 10 –3 m) 2 )(11ºC) = 3.4 N. Then, 
F/F = – 0.034, 
f/f = – 0.017, and finally, 
f = –4.2 Hz, or the pitch falls. This also explains the constant the constant tuning in the string sections of symphonic orchestras.
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