Answer
See answers.
Work Step by Step
a. The fundamental frequency of the vibrating string is given by $f_1=\frac{v}{2\mathcal{l}}$. For this situation, is constant (because the string tension has not changed). The frequency is inversely proportional to the length. To raise the frequency by a factor of $\frac{440\;Hz}{330\;Hz}=4/3$, the length should be reduced by the same factor.
$$\mathcal{l}_{new}=\frac{\mathcal{l}_{old}}{4/3}=0.75(0.68m)=0.51m$$
Therefore, the string should be fretted at a distance of 0.68m-0.51m = 0.17m from the end.
b. The string is fixed at both ends, and is vibrating at its fundamental frequency. The wavelength is twice the length of the string.
$$\lambda=2\mathcal{l}=2(0.51\;m)=1.02\;m$$
c. The frequency of the emitted sound is also 440 Hz. The wavelength of the sound is calculated as follows, using the speed of sound as a function of temperature that is the first equation in Chapter 12.
$$\lambda=\frac{v}{f}=\frac{331+(0.60)(22)}{440Hz}=0.78\;m$$
