A wire fixed at both its ends is found to vibrate at resonant...

For a string fixed at both ends, the resonant frequencies are given by $f_n = \dfrac{nv}{2L}$, where $n$ is an integer representing the harmonic number, $v$ is the wave speed, and $L$ is the length of the string.

Assuming the given frequencies are consecutive resonant frequencies, their difference gives the fundamental frequency $f_1$:

$f_1 = f_{n+1} - f_n$

$f_1 = 320 \text{ Hz} - 240 \text{ Hz} = 80 \text{ Hz}$

The fundamental frequency is also given by the formula:

$f_1 = \dfrac{v}{2L}$

Substituting the given values ($f_1 = 80 \text{ Hz}$ and $v = 40 \text{ m s}^{-1}$):

$80 = \dfrac{40}{2L}$

$160L = 40$

$L = \dfrac{40}{160} \text{ m}$

$L = 0.25 \text{ m}$

Converting the length to centimeters:

$L = 0.25 \times 100 \text{ cm} = 25 \text{ cm}$