1. For a sonometer wire vibrating in fundamental mode,

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

where:

• $f$ = fundamental frequency
• $L$ = resonating length
• $T$ = tension
• $\mu$ = linear mass density

2. Since the tension and the wire are the same, $\sqrt{T/\mu}$ is constant. So,

$f \propto \frac{1}{L}$

Hence,

$f_1 L_1 = f_2 L_2$

3. Given:

$f_1 = 400\,\text{Hz}, \quad L_1 = 90\,\text{cm}, \quad f_2 = 600\,\text{Hz}$

So,

$L_2 = \frac{f_1 L_1}{f_2} = \frac{400 \times 90}{600}$

$L_2 = 60\,\text{cm}$

4. Therefore, the required resonating length is

$\boxed{60}$

Comparison with stored correct answer:

• Derived answer = $60$
• Stored correct answer = $60$
• They match.