Out of Tune. The B-string of a guitar is made of steel (density 7800 kg/m3), is 63.5 cm long, and has diameter 0.406 mm. The fundamental frequency is ?f = 247.0 Hz. (a) Find the string tension. (b) If the tension ?F is changed by a small amount ??F?, the frequency ?f changes by a small amount ?f ? ?. Show that (c) The siring is tuned to a fundamental frequency of 247.0 Hz when its temperature is 18.5°C. Strenuous playing can make the temperature of the string rise, changing its vibration frequency. Find ??f if the temperature of the siring rises to 29.5°C. The steel string has a Young’s modulus of 2.00 × 1011 Pa and a coefficient of linear expansion of 1.20 × 10–5 (C°)–1. Assume that the temperature of the body of the guitar remains constant. Will the vibration frequency rise or full?
Solution 87P Step 1 The density of the material of the string is = 7800 kg/m . d2 (0.406×1m)2 3 So the linear density of the string is = A =4) = (7800 kg)( 4 ) = 1.01 × 10 kg/m Since the string is vibrating with its fundamental frequency, the wavelength of the string is = 2L = 2(0.635 m) = 1.27 m And the frequency is f = 247 Hz Hence the speed of the sound is c = f = (1.27 m)(247 Hz) = 313.7 m/s Now the speed of the sound is given by v = F F 313.7 m/s = 1.01×10kg/m F = 99.4 N So the tension of the string is 99.4 N. Step 2 (b) The speed of sound is given by v = F The speed can also be written as v = f Using the above equation we can write that F f = 2 2 F f = …..(1) Now can write that 2 dF 2 fdf = ….(2) Dividing equation (2) with equation (1) we can write that df dF 2 f = F df = dF (proved) f 2F