Question

which is 1.5 mm in radius is tuned to its at a Hawaiian luau on a warm day (32°C). 2. A ukelele has steel strings which are 0.35 m long. The string which is 1.5 mm in radius Tirst harmonic occurs at 440. Hz). The ukulele is being played at a Hawaiian luau on a warm Sketch the string vibrating at its first harmonic. Find a) the velocity of the waves on the string b) the frequency of the second harmonic that could be achieved on this string c) the tension in the string and whether this tension is likely to break the string (show supporting info for your answer) d) the frequency heard by a hula dancer when she is moving at 2.5 m/s toward the ukulele player

Answer 1

The string vibrating in its first harmonic. < --- In this case, the wavelength is: 2 = 2L a) The speed of the waves is giving by: v = 2f = 2Lf = 2(0.35)(440) = 308 b) The frequency of the second harmonic is: f2 = 2f1 = 2(440) = 880 Hz c) Remember that: v= - Then, the force is: F = uv2 The liner mass of the steel string is: т u= ī The mass is: m = pV = pal Then: u = PA = pur2 Then: F = par2y2 Consider the density of the steel is 7800 kg/m3 F = 78001(1.5 x 10-3)2(308)2 = 5230 N The stress is: F 5230 E = A Ā Al 1(1.5 x 10-3)2 = 7.40 X 108Pa For steel the rupture stress is about 4 x 108 Pa so the string will break d) The speed of sound at 32°C is: T 32 Vs = 331 |1 +372 = 331 1 +222 = 350 – 273 Using the Doppler effect equation: f(vs + v.) 440(350 +2.5) f = flus TV =- = 443 Hz vs 350