A string fixed at both ends vibrates in a resonant mode, having a...

Let the length of the string be $L$.

For a string fixed at both ends vibrating in the $n$-th harmonic, the separation between consecutive nodes is $\dfrac{\lambda}{2} = \dfrac{L}{n}$.

Given that for the $n$-th harmonic, the separation is $2.0 \text{ cm}$, we have:

$L = n \times 2.0$

For the next higher resonant mode, the string vibrates in the $(n+1)$-th harmonic. The separation between consecutive nodes becomes $1.6 \text{ cm}$, so:

$L = (n+1) \times 1.6$

Equating the two expressions for $L$:

$2.0 n = 1.6 (n+1)$

$2.0 n = 1.6 n + 1.6$

$0.4 n = 1.6$

$n = 4$

Substituting the value of $n$ into the first equation:

$L = 4 \times 2.0 = 8.0 \text{ cm}$