A tuning fork having a frequency of 512 Hz is vibrated alongside a...

Let the frequency of the tuning fork be $f_{t} = 512 \text{ Hz}$.

The beat frequency is $n = 6 \text{ Hz}$.

The frequency of the sonometer wire $f_{w}$ is given by $f_{w} = f_{t} \pm n$.

$f_{w} = 512 + 6 = 518 \text{ Hz}$ or $f_{w} = 512 - 6 = 506 \text{ Hz}$.

The frequency of a sonometer wire is directly proportional to the square root of tension ($f \propto \sqrt{T}$).

When the tension in the wire is slightly increased, its frequency $f_{w}$ increases.

If the original frequency was $518 \text{ Hz}$, increasing the tension would increase the frequency, resulting in a new beat frequency greater than $6 \text{ Hz}$.

If the original frequency was $506 \text{ Hz}$, increasing the tension would increase the frequency towards $512 \text{ Hz}$, resulting in a new beat frequency less than $6 \text{ Hz}$.

Since the beat frequency decreases, the original frequency of the wire must be $506 \text{ Hz}$.