Wave Motion> Two tuning forks with natural frequencies...

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 , 6 Years ago

Grade 12th pass

2 Answers

Radha Shukla

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

Last Activity: 6 Years ago

Khimraj

Hello Student,

Please find the answer to your question

The apparent frequency from tuning fork T₁ as heard by the observer will be

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

Last Activity: 6 Years ago

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Wave Motion> Two tuning forks with natural frequencies...

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 , 6 Years ago

Radha Shukla

Khimraj

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