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Step-by-Step Solution
Step 1: Recall the formula for fundamental frequency of a vibrating string
The fundamental frequency f of a stretched string is given by
$f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,
where
$l$ is the length of the string,
$T$ is the tension in the string, and
$\mu$ is the linear mass density of the string.
Step 2: Observe that length and linear mass density are identical for both strings
Since the question states that both strings X and Z are identical (same material, same dimensions), their length $l$ and linear mass density $\mu$ are the same. Therefore, the only difference affecting their fundamental frequencies is the tension $T$.
Step 3: Use the proportional relationship
From
$f \propto \sqrt{T}$,
we see that
$\frac{f_X}{f_Z} = \sqrt{\frac{T_X}{T_Z}}$,
where $f_X$ and $f_Z$ are the fundamental frequencies of strings X and Z, respectively, and $T_X$ and $T_Z$ are the corresponding tensions.
Step 4: Substitute the known frequencies
We are given
$f_X = 450\,\text{Hz}$
and
$f_Z = 300\,\text{Hz}$.
So,
$\frac{450}{300} = \sqrt{\frac{T_X}{T_Z}}$.
Step 5: Simplify to find the ratio of tensions
Simplifying the left side,
$\frac{450}{300} = \frac{3}{2}$.
Hence,
$\frac{3}{2} = \sqrt{\frac{T_X}{T_Z}}$.
Squaring both sides, we get
$\frac{T_X}{T_Z} = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25.$
Final Answer
The ratio of the tensions is
$\frac{T_X}{T_Z} = 2.25.$
