A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one receded with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v_(0)= 1400Hz and the velocity of sound in air is 350m/s. The speed of each tuning fork is close to

A stationary observer receives sound from two identical tuning forks,

1.	A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one receded with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v_(0)= 1400Hz and the velocity of sound in air is 350m/s. The speed of each tuning fork is close to
Answer» `(1)/(4)` 4 2 `(1)/(2)` Solution : `f_(0) ((c )/(c-v)) - f_(0) ((c )/(c+ v)) =2` [Velocity of sound C= 350 m/s] `1400 ((c^(2) + cv- c^(2) + cv)/(c^(2)- v^(2))) =2` `700 xx 2cv = c^(2) -v^(2)` `v^(2) + 1400cv - c^(2) = 00` By TAKING C= 350 `v^(2) + 1400 xx 350 v- [350]^(2)= 0` `:. v^(2) + 490000 v- 122500= 0` Which is QUADRATIC equation `:. a=1, b= 490000c, = -122500` `:. sqrtDelta = sqrt(b^(2) - 4ac)` `= sqrt((490000)^(2)- 4 xx 1 xx (-122500))` `=sqrt(2401 xx 10^(8) + 490000)` `= sqrt(24010049 xx 10^(4))` `= 4900.00499 xx 10^(2)` =490000.5 Now `v= (-b +- sqrtDelta)/(2a)` `=(-490000 +- 490000.5)/(2 xx 1)` `= (-490000 + 490000.5)/(2) or (-490000-490000.5)/(2)` `= (0.5)/(2) = (1)/(4)`= This is impossible Because `(v LT c)` velocity v should be less than c.