The fundamental frequency of a string (in this case, a sonometer wire) when it is vibrating, is inversely proportional to its length, provided the tension in the string and the linear mass density (mass per unit length) remain constant. This relationship is given by the formula:

$$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$

where:

• $f$ is the frequency of the string,
• $L$ is the length of the string,
• $T$ is the tension in the string, and
• $\mu$ is the linear mass density of the string.

Given that the sonometer wire has a resonating length of 90 cm (0.9 m) with a fundamental frequency of $400 \mathrm{Hz}$, we can set up our first equation but note that we are comparing two states of the same string under the same tension and hence can eliminate the tension and density terms for comparison purposes:

$$400 = \frac{1}{2 \times 0.9} \sqrt{\frac{T}{\mu}}$$

For the second scenario where the fundamental frequency is $600 \mathrm{Hz}$, we are asked to find the new length $L_2$. We can set up the equation in a similar manner:

$$600 = \frac{1}{2L_2} \sqrt{\frac{T}{\mu}}$$

Since the tension $T$ and the linear density $\mu$ are constants, and they do not change between the two states, we can set up a proportion between the two states by dividing the second equation by the first, which yields:

$$\frac{600}{400} = \frac{\frac{1}{2L_2}}{\frac{1}{2 \times 0.9}}$$

Simplifying this equation gives:

$$\frac{600}{400} = \frac{0.9}{L_2}$$

or

$$\frac{3}{2} = \frac{0.9}{L_2}$$

Solving for $L_2$ gives:

$$L_2 = \frac{0.9 \times 2}{3}$$

Calculating the value:

$$L_2 = \frac{1.8}{3} = 0.6 \hspace{1mm} \text{meters}$$

Converting meters to centimeters (since 1 meter = 100 centimeters), we find:

$$L_2 = 0.6 \times 100 = 60 \hspace{1mm} \text{centimeters}$$

Therefore, the resonating length of the wire with a fundamental frequency of $600 \mathrm{Hz}$ under the same tension is 60 cm.