When two tuning forks are sounded together, 6 beats per second are heard. One of the fork is in unison with 0.70 m length of sonometer wire and another fork is in unison with 0.69 m length of the same sonometer wire. The frequencies of the two tuning forks are

Step-by-step solution

When two tuning forks produce 6 beats per second, the difference in their frequencies is 6 Hz: |f₁ - f₂| = 6 . The frequency of a vibrating sonometer wire is inversely proportional to its length, f 1/L , so fL remains constant for constant tension and linear density. Given L₁ = 0.70 m and L₂ = 0.69 m , the frequencies satisfy f₁ L₁ = f₂ L₂ . Since L₁ > L₂ , it follows that f₁ Substitute f₁ = f₂ L₂ / L₁ into the beat equation: f₂ - (0.69 / 0.70) f₂ = 6 . Simplify to f₂ (1 - 0.69 / 0.70) = 6 , yielding f₂ (0.01 / 0.7