A string fixed at both the ends, forms a standing wave with node...

The distance between two consecutive nodes in a standing wave is $\dfrac{\lambda}{2}$.

Given that the node separation is $5\ \text{cm}$, we have:

$\dfrac{\lambda}{2} = 5\ \text{cm} = 0.05\ \text{m}$

$\lambda = 0.1\ \text{m}$

The velocity of the wave is given as $v = 2\ \text{m/s}$.

The frequency of vibration $f$ is related to the wave velocity and wavelength by the formula:

$f = \dfrac{v}{\lambda}$

Substituting the values:

$f = \dfrac{2}{0.1} = 20\ \text{Hz}$

Answer: $20\ \text{Hz}$