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Two tuning forks with natural frequencies v1 and v2 respectively, are struck at the same time with equal force. The intensity, of the resulting sound, waxes and wanes with time period of 1.5 seconds while the frequency of the sound is 256 Hz. Hence, v1 and v2 . respectively, are

Akib hassan , 5 Years ago
Grade 12th pass
anser 2 Answers
Radha Shukla

Last Activity: 5 Years ago

Intensity is directly proportional to square root of frequency as force is same so amplitude will also be same... here intensity will vary with square of frequency 

Khimraj

Last Activity: 5 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 2
∴ v = 3 x 340 x 340/340 x 340 x 2 = 1.5 m/s

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