A tuning fork resonates with a sonometer wire of length $$1$$ m stretched with a tension of $$6$$ N. When the tension in the wire is changed to $$54$$ N, the same tuning fork produces $$12$$ beats per second with it. The frequency of the tuning fork is _______ Hz.

A tuning fork resonates with a wire under tension 6 N. Under tension 54 N, it produces 12 beats/s. Find the tuning fork frequency.

The fundamental frequency of a vibrating string is given by $$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $$, where $$L$$ is length, $$T$$ is tension, and $$\mu$$ is mass per unit length. Since $$L$$ and $$\mu$$ are constant for the same wire, $$f \propto \sqrt{T}$$.

Let the frequency with tension 6 N be $$f_1$$ and with 54 N be $$f_2$$. Then $$ \frac{f_2}{f_1} = \sqrt{\frac{54}{6}} = \sqrt{9} = 3, $$ so $$f_2 = 3f_1$$.

The tuning fork resonates with the wire at tension 6 N, so its frequency $$f_{\text{fork}}$$ equals $$f_1$$.

Since beats occur when two frequencies are close but not equal, the beat frequency is the absolute difference between the wire’s frequency under 54 N tension and the fork’s frequency: $$ |f_2 - f_{\text{fork}}| = 12. $$ Substituting $$f_2 = 3f_1$$ and $$f_{\text{fork}} = f_1$$ gives $$ |3f_1 - f_1| = 2f_1 = 12, $$ hence $$ f_1 = 6 \text{ Hz}. $$

The answer is 6 Hz.