Question

The fequency of tuning fork A and B are 2% more and 2% less than the frequency of tuning fork ​

Answer 1

your question is incomplete. A complete question should be ----> The frequency of tuning fork A is 2% more than the frequency of a standard fork. Frequency of tuning fork B is 3% less than the frequency of the standard fork. If 6 beats per second are heard when the two forks A and B are excited, then frequency of A is……….Hz.

solution : Let the frequency of standard tuning fork is x

then, frequency of tuning fork A, $f_A$ = x + 2% of x

= x + 2x/100 = 102x/100

Similarly, frequency of tuning fork B, $f_B$ = x - 3% of x

= x - 3x/100 = 97x/100

given, $f_A-f_B$ = 6 beats/sec

or, 102x/100 - 97x/100 = 6

or, 5x/100 = 6

or, x/20 = 6

or, x = 120 Hz

so, frequency of tuning fork A = 102 × 120/100 = 122.4 Hz