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Two tuning forks with natural frequencies of 340 Hz each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards him at the same speed of the tuning fork.

Simran Bhatia , 10 Years ago
Grade 11
anser 1 Answers
Aditi Chauhan

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
The apparent frequency from tuning fork T1 as heard by the observer will be
236-1628_6.png
v1 = c/c – v x v . . . . . . . . . . . . . . . . . . . . . . . (i)
where c = velocity of sound
v = velocity of turning fork
The apparent frequency from tuning fork T2 as heard by the observer will be
v2 = c/c + v x v . . . . . . . . . . . . . . . . . (ii)
Given v1 – v2 = 3
∴ c x v [1/c – v – 1/ c + v] = 3 or, 3 = c x v x 2v/c2 – v2
Since, v < < c ∴ 3 = c x v x 2v/c2
∴ v = 3 x 340 x 340/340 x 340 x 2 = 1.5 m/s

Thanks
Aditi Chauhan
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