P18_033.PDF

10 − 6 kg) / (0 . 220 m) (405 m / s) 2 = 596 N.

(d) The frequency of the sound wave in air is the same as the frequency of oscillation of the string.

The wavelength is diﬀerent because the wave speed is diﬀerent. If v a is the speed of sound in air

the wavelength in air is λ a = v a /f = (343 m / s) / (920 Hz) = 0 . 373 m.

33. (a) When the string (ﬁxed at both ends) is vibrating at its lowest resonant frequency, exactly one-

half of a wavelength ﬁts between the ends. Thus, λ =2 L .Weobin v = fλ =2 Lf =

2(0 . 220 m)(920 Hz) = 405 m / s.

(b) The wave speed is given by v = τ/µ ,where τ is the tension in the string and µ is the linear

mass density of the string. If M is the mass of the (uniform) string, then µ = M/L .Thus

τ = µv 2 =( M/L ) v 2 = (800