Answer
See answers.
Work Step by Step
a. The tuner hears a beat frequency of 3 beats per 2.5 seconds, or 1.2 Hz. Assume that the strings are the same length and that they have the same mass density.
a. The other string has a frequency that differs by 1.2 Hz from 220.0 Hz, so its frequency is either 218.8 Hz, or 221.2 Hz.
b. Note how the frequency and tension are related.
$$f=\frac{v}{2\mathcal{l}}=\frac{1}{2\mathcal{l}}\sqrt{\frac{F_{tension}}{\mu}}$$
The higher the tension, the higher the frequency. We relate the new string tension to the old one.
$$ F_{tension2}= F_{tension1}(\frac{f_2}{f_1})^2$$
To change 218.8 Hz to 220.0 Hz, calculate the new tension.
$$ F_{tension2}= F_{tension1}(\frac{220.0Hz}{218.8Hz})^2= F_{tension1}(1.011)$$
A 1.1 percent increase in tension on the 218.8 Hz string will bring the 2 strings in tune.
To change 221.2 Hz to 220.0 Hz, calculate the new tension.
$$ F_{tension2}= F_{tension1}(\frac{220.0Hz}{221.1Hz})^2= F_{tension1}(0.9892)$$
A 1.1 percent decrease in tension on the 221.2 Hz string will bring the 2 strings in tune.
