A cylindrical metal tube is open at both ends and has a length of...

For a tube open at both ends, the resonant frequencies are given by $f_n = \dfrac{nv}{2L}$, where $n = 1, 2, 3, \dots$

Given $L = 50 \text{ cm} = 0.5 \text{ m}$ and $v = 340 \text{ m s}^{-1}$, the fundamental frequency is:

$f_1 = \dfrac{340}{2 \times 0.5} = 340 \text{ Hz}$

The possible resonant frequencies are integral multiples of the fundamental frequency:

$f_n = n \times 340 \text{ Hz}$

Calculating the frequencies for different values of $n$:

For $n = 3$, $f_3 = 3 \times 340 = 1020 \text{ Hz}$

For $n = 4$, $f_4 = 4 \times 340 = 1360 \text{ Hz}$

For $n = 5$, $f_5 = 5 \times 340 = 1700 \text{ Hz}$

For $n = 6$, $f_6 = 6 \times 340 = 2040 \text{ Hz}$

The frequencies that lie strictly between $1000 \text{ Hz}$ and $2000 \text{ Hz}$ are $1020 \text{ Hz}$, $1360 \text{ Hz}$, and $1700 \text{ Hz}$.