Two tuning forks, vibrating simultaneously, produce 6 beats per second. The first of them has a frequency of 312 Hz . Some amount of wax is added to one arm of the second tuning fork , the number of beats per second reduces to 3 . Find out the frequency of this second tuning fork . Is it possible to increase the beat frequency to 6 per second by adding some more wax to the second tuning fork ?

Answer

Initially, the number of beats per second is 6 . So the frequency of the 2nd fork is either 312 + 6 = 318 Hz or 312 - 6 = 306 Hz . After putting the wax, the beat frequency is 3 per second. So the changed frequency of the 2nd fork is, either 312 + 3 = 315 Hz or 312 - 3 = 309 Hz . But the addition of wax reduces the frequency of the 2nd fork ; so it cannot change from 306 Hz to 309 Hz ; the actual change is from 318 Hz to 315 Hz . This means that the actual frequency of the 2nd tuning fork is 318 Hz .
By adding a sufficient amount of wax on the 2nd fork, its frequency can be reduced from 318 Hz to 306 Hz . Then again the beat frequency would be 312 - 306 = 6 per second .