## College Physics (4th Edition)

(a) The speed of the transverse waves in the string is $1350~m/s$ (b) The tension is $45.6~N$ (c) The frequency in the air is $450.0~Hz$ The wavelength in the air is $0.756~m$
(a) We can find the speed of the transverse waves in the string: $v = \lambda~f$ $v = 2L~f$ $v = (2)(1.50~m)(450.0~Hz)$ $v = 1350~m/s$ The speed of the transverse waves in the string is $1350~m/s$ (b) We can find the tension: $\sqrt{\frac{F}{\mu}} = v$ $\frac{F}{\mu} = v^2$ $F = \mu~v^2$ $F = (25.0\times 10^{-6}~m)(1350~m/s)^2$ $F = 45.6~N$ The tension is $45.6~N$ (c) The frequency in the air is the same as the frequency in the string. Therefore, the frequency in the air is $450.0~Hz$ We can find the wavelength in the air: $\lambda = \frac{v}{f}$ $\lambda = \frac{340~m/s}{450.0~Hz}$ $\lambda = 0.756~m$ The wavelength in the air is $0.756~m$