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Asked by saineerajbudumuru1218 | 30 Jun, 2020, 04:54: PM

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A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes 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 v0=1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to -

O n e space t u n i n g space f o r k space a p p r o a c h e s space a n d space o t h e r space r e c e d e s space T h u s comma space f subscript 1 space equals space open parentheses fraction numerator C over denominator C minus v end fraction close parentheses f subscript 0 space f subscript 2 space equals space open parentheses fraction numerator C over denominator C plus v end fraction close parentheses f subscript 0 space B e a t space f r e q u e n c y space equals space f subscript 1 space minus space f subscript 2 space equals space 2 T h u s comma space space c f subscript 0 space open parentheses fraction numerator 1 over denominator c minus v end fraction minus fraction numerator 1 over denominator c plus v end fraction close parentheses space equals space 2 equals c f subscript 0 space open parentheses fraction numerator c plus v minus c plus v over denominator c squared space minus space v squared end fraction close parentheses space equals space fraction numerator 2 c f subscript 0 v over denominator c squared space minus space v squared end fraction almost equal to fraction numerator 2 f subscript 0 v over denominator c end fraction space equals 2 space rightwards double arrow fraction numerator 2 cross times 1400 cross times v over denominator 350 end fraction equals 2 v space equals space 1 fourth space m divided by s

Thus,correct option is (2)

Answered by Shiwani Sawant | 02 Jul, 2020, 10:44: PM

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