Step 1: Understanding the Concept:
The beat frequency between two tuning forks is the absolute difference between their individual frequencies.
A tuning fork in unison with a sonometer wire has the same fundamental frequency as that segment of the wire.
For a stretched string under constant tension, the fundamental frequency is inversely proportional to its vibrating length.
Step 2: Key Formula or Approach:
Beat frequency: $n = |f_1 - f_2| = 6 \text{ Hz}$.
Sonometer frequency relation: $f \propto \frac{1}{L}$.
Therefore, $\frac{f_1}{f_2} = \frac{L_2}{L_1}$.
We have two equations and two unknowns to solve for $f_1$ and $f_2$.
Step 3: Detailed Explanation:
Let the frequencies of the two tuning forks be $f_1$ and $f_2$.
Let $f_1$ be in unison with length $L_1 = 0.70 \text{ m}$.
Let $f_2$ be in unison with length $L_2 = 0.69 \text{ m}$.
Since length $L_1>L_2$, its corresponding frequency must be lower: $f_1<f_2$.
From the beat frequency information:
\[ f_2 - f_1 = 6 \quad \text{--- (Equation 1)} \]
From the law of length for stretched strings:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} = \frac{0.69}{0.70} = \frac{69}{70} \]
Let $f_1 = 69k$ and $f_2 = 70k$, where $k$ is a constant multiplier.
Substitute these into Equation 1:
\[ 70k - 69k = 6 \]
\[ k = 6 \]
Now calculate the actual frequencies:
\[ f_1 = 69 \times 6 = 414 \text{ Hz} \]
\[ f_2 = 70 \times 6 = 420 \text{ Hz} \]
The frequencies are $414 \text{ Hz}$ and $420 \text{ Hz}$.
Step 4: Final Answer:
The frequencies are 414 Hz, 420 Hz.