A tuning fork arrangement (pah) produces 4 beats/sec with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The frequency of unknown fork is

 Frequency of unknown fork  =$=(288\pm 4)$cps When unknown fork is loaded with wax, its frequency decreases. Given that beat frequency decreases to z
beats/sec. This shows that the frequency of unknown fork is greater than 288 Hz . So, the frequency of unknown fork must be $(288+4)=292\mathrm{cps}$