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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 -

1 4 m / s

Option 2 -

1 m / s

Option 3 -

1 2 m / s

Option 4 -

1 8 m / s

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 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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An observer moves towards a stationary source of sound with a velocity one-fourth of the velocity of sound. What is the percentage increase in the apparent frequency?
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Vishal Baghel

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⇒ f ' f = 5 4

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The driver of bus approaching a big wall notices that the frequency of his bus’s horn changes from 420 Hz to 490 Hz when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is 330 ms⁻¹ :
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Vishal Baghel

f? = 420 Hz
f? = (V? + V)/ (V? - V) f? = 490
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(330 + V)6 = 7 (330 – V)

13 V = 330
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A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from 400 Hz to 400 Hz , when it gets reflected from the wall. If the speed of sound in air is 345 m/s , then the speed of the car is:
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Vishal Baghel

f? = frequency heard by wall = f_s (v_s/ (v_s - v_c)
f? = frequency heard by driver after reflection from wall


f? = (v_s + v_c)/v_s)f? = (v_s + v_c)/ (v_s - v_c)f?
f? /f? = (v_s - v_c)/ (v_s + v_c)
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