A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much les than the speed of sound). The observers hears 2 beats/sec. The oscillation frequency of each tuning fork is v 0 = 1400 H z  and the velocity of sound in air is 350 m / s . The speed of each tuning fork is close to:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mo>/</mo> <mi mathvariant="normal">s</mi> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>1</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mo>/</mo> <mi mathvariant="normal">s</mi> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mo>/</mo> <mi mathvariant="normal">s</mi> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </mfrac> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mo>/</mo> <mi mathvariant="normal">s</mi> </math> </span></p>
2 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months ago
Correct Option - 1
Detailed Solution:

  f 1 = f o v v - v s f 1 - f 2 = 2

f 2 = f o v v + v s f 0 v v + v s - v - v s v 2 - v s 2 = 2 ; 2 f 0 v v s v 2 - v s 2 = 2

? v s ? v v s = 1 4 m / s

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Physics Waves 2021

Physics Waves 2021

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