A specific guitar wire has a length of 30.0 cm and vibrates at a...

The fundamental frequency of a stretched string is given by the formula:

$f = \dfrac{v}{2L}$

Since the tension and the linear mass density of the guitar wire remain constant, the wave speed $v$ is constant. Therefore, the fundamental frequency is inversely proportional to the vibrating length of the string:

$f \propto \dfrac{1}{L} \Rightarrow f_0 L_0 = f_1 L_1 = f_2 L_2 = \dots$

Given the original length $L_0 = 30.0 \text{ cm}$ and the original frequency $f_0 = 196 \text{ Hz}$, we can find the required vibrating lengths for the subsequent frequencies.

For $f_1 = 220 \text{ Hz}$:
$L_1 = \dfrac{f_0 L_0}{f_1} = \dfrac{196 \times 30.0}{220} = 26.72 \text{ cm} \approx 26.7 \text{ cm}$

For $f_2 = 247 \text{ Hz}$:
$L_2 = \dfrac{f_0 L_0}{f_2} = \dfrac{196 \times 30.0}{247} = 23.80 \text{ cm} \approx 23.8 \text{ cm}$

For $f_3 = 262 \text{ Hz}$:
$L_3 = \dfrac{f_0 L_0}{f_3} = \dfrac{196 \times 30.0}{262} = 22.44 \text{ cm} \approx 22.4 \text{ cm}$

For $f_4 = 294 \text{ Hz}$:
$L_4 = \dfrac{f_0 L_0}{f_4} = \dfrac{196 \times 30.0}{294} = 20.0 \text{ cm}$

These lengths represent the distances from the bridge (the lower end of the string) where the finger must be positioned to produce the respective notes. Thus, the required distances are $26.7 \text{ cm}$, $23.8 \text{ cm}$, $22.4 \text{ cm}$, and $20.0 \text{ cm}$.

Answer: $26.7 \text{ cm}$, $23.8 \text{ cm}$, $22.4 \text{ cm}$ and $20.0 \text{ cm}$