1. For a stretched string, the fundamental frequency is

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

where:

• $L$ = length of string
• $T$ = tension
• $\mu$ = linear mass density

2. Since the strings are identical and made of the same material, they have the same $L$ and same $\mu$.

So,

$f \propto \sqrt{T}$

3. Therefore,

$\frac{f_X}{f_Z} = \sqrt{\frac{T_X}{T_Z}}$

Given:

$f_X = 450\text{ Hz}, \qquad f_Z = 300\text{ Hz}$

Thus,

$\frac{450}{300} = \sqrt{\frac{T_X}{T_Z}}$

$\frac{3}{2} = \sqrt{\frac{T_X}{T_Z}}$

4. Squaring both sides,

$\frac{T_X}{T_Z} = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25$

5. Check options:

• A: $2.25$ ✅
• B: $0.44$ ❌
• C: $1.25$ ❌
• D: $1.5$ ❌

Hence, the correct answer is A.