1. Use the harmonic frequency relation for a stretched string

For a string of length $L$ fixed at both ends,

$f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$

where:

• $f_n$ = $n$th harmonic frequency
• $T$ = tension
• $\mu$ = linear mass density

2. Use the difference of consecutive harmonics

Given:

• $f_n = 400\,\text{Hz}$
• $f_{n+1} = 450\,\text{Hz}$

So,

$f_{n+1} - f_n = 450 - 400 = 50\,\text{Hz}$

But for a stretched string,

$f_{n+1} - f_n = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Given $L = 30\,\text{cm} = 0.30\,\text{m}$, hence

$50 = \frac{1}{2(0.30)}\sqrt{\frac{2700}{\mu}}$

3. Solve for $\mu$

First simplify:

$2L = 0.60$

So,

$50 = \frac{1}{0.60}\sqrt{\frac{2700}{\mu}}$

$\sqrt{\frac{2700}{\mu}} = 50 \times 0.60 = 30$

Squaring both sides,

$\frac{2700}{\mu} = 900$

$\mu = \frac{2700}{900} = 3\,\text{kg/m}$

4. Final answer

$\boxed{3}$

So the linear mass density is $3\,\text{kg/m}$.