Question

frequency of tuning fork A is 256 hertz it produces 4 beats per second with tuning fork B when wax was applied at tuning fork B then 6 beats per second are heard .By reducing little amount of wax 4 beats per second are heard .Frequency of B is?​

Answer 1

Answer:

256 - nB = 4 eq 1

nB - 256 = 4. eq 2

nB is Decrease

number of Beat increase

eq 1 is right

256- nB = 4

nB = 252Hz