Answer
(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 $0.756~m$
Work Step by Step
(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 $0.756~m$
