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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

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 

Grade:12th pass

2 Answers

Radha Shukla
178 Points
2 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
3007 Points
2 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
v= 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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