Question

A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s. [ Use g = 10 m/s2
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Answer 1

as approching, apparent frequency increases for the velocity of the apporoching tuningfor f1 be +v the others -v

f1-f2 = 680*340( 1/(340-v)- 1/(340+v)) = 2      =========       

also f1 = 680*340/(340-v)       ===========     you have 2 equations and 2 variables now solve by componento dividento easily 

Answer 2

$\underline\bold{Answer}$

${( \frac{v}{v - vs} ) - ( \frac{v}{v + vs} )}fo \: = 2hz \\ \\ vs \: = 0.5m {s}^{ - 1}$

$\underline\bold{Thanks}$