A vibrating string of certain length $$\ell $$ under a tension $$T$$ resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $$75\,cm$$  inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency $$n.$$ Now when the tension of the string is slightly increased the number of beats reduces 2 per second. Assuming the velocity of sound in air to be $$340\,m/s,$$  the frequency $$n$$ of the tuning fork in $$Hz$$  is

Solution :
The frequency $$\left( \nu \right)$$  produced by the air column is given by
$$\eqalign{ & \lambda \times \nu = \nu \cr & \Rightarrow \,\,\nu = \frac{\nu }{\lambda } \cr & {\text{Also, }}\frac{{3\lambda }}{4} = \ell = 75cm = 0.75m \cr & \therefore \,\,\lambda = \frac{{4 \times 0.75}}{3} \cr & \Rightarrow \,\nu = \frac{{340 \times 3}}{{4 \times 0.75}} \cr & = 340Hz \cr} $$
$$\therefore $$ The frequency of vibrating string = 340. Since this string produces 4 beats/sec with a tuning fork of frequency $$n$$ therefore $$n = 340 + 4$$   or $$n = 340 - 4.$$   With increase in tension, the frequency produced by string increases. As the beats/sec decreases therefore $$n = 340 + 4 = 344\,Hz.$$