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Step-by-Step Solution
Step 1: Understand the problem
A wire with mass per unit length $ \mu = 6.0 \times 10^{-3} \, \text{kg m}^{-1} $ is under a tension $ T = 540 \, \text{N} $. The wire resonates at two consecutive frequencies, 420 Hz and 490 Hz. We need to determine its length $L$.
Step 2: Identify the fundamental frequency
For consecutive resonant frequencies of a string, the difference between those frequencies is equal to the fundamental frequency (i.e., the first harmonic). Thus, the fundamental frequency $ f_1 $ is:
$ f_1 = 490 \, \text{Hz} - 420 \, \text{Hz} = 70 \, \text{Hz} $
Step 3: Apply the formula for the fundamental frequency of a stretched string
The fundamental frequency of a string fixed at both ends is given by:
$ f_1 = \frac{1}{2 L} \sqrt{\frac{T}{\mu}}
$
where
$ L $ = length of the wire
$ T $ = tension in the wire
$ \mu $ = mass per unit length
Step 4: Substitute known values and solve for the length $ L $
Given $ f_1 = 70 \, \text{Hz}, \, T = 540 \, \text{N}, \, \mu = 6.0 \times 10^{-3} \, \text{kg m}^{-1} $,
substitute these into the formula:
$ 70 = \frac{1}{2L} \sqrt{\frac{540}{6.0 \times 10^{-3}}}
$
$ \sqrt{\frac{540}{6.0 \times 10^{-3}}} = \sqrt{9.0 \times 10^{4}} = 300
$
So the equation becomes:
$ 70 = \frac{300}{2L}
$
$ 70 = \frac{300}{2L} \quad \Rightarrow \quad 2L = \frac{300}{70} \quad \Rightarrow \quad 2L = \frac{30}{7}
$
$ L = \frac{30}{14} = \frac{15}{7} \approx 2.14 \,\text{m}
Step 5: Final answer
The length of the wire $ L \approx 2.1 \,\text{m}$ (rounding to one decimal place).
