Question

Physics
Check Your Understanding
Frequency of tuning fork A is 256 Hz. It produces
four beats/sec with tuning fork B. When wax is
applied at tuning fork B then 6 beats/sec are heard.
By reducing little amount of wax 4 beats/sec are
heard. Frequency of B is :
(1) 250 Hz
(2) 260 Hz
(3) 252 Hz
(4) 256 Hz
please give a detailed answer​

Answer 1

let's call frequency 'n'

n(A) is 256 Hz and gives 4 beats/sec with n(B).

so n(B) has to be either 256+4= 260

or 256-4=252

now wax is applied which reduces frequency of B. this new frequency of B gives 6 beats.

Which means that the difference between the nA and nB has increased.

Apply logic... If nB was 260 Hz and it reduces... it will come closer to nA that is 256 Hz. and beats should decrease... but that's not the case.

so answer is n(B)= 252 Hz

you can verify: 252 reduced to 250.. gives 6 beats : 256-250

and 256-252= 4 beats..

this is an easy method by which I solve these kind of problems..

there are many more mathematical techniques but this one is fastest

hope it helps!