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The sound emitted from the siren of an ambulance has a frequency of 1500 Hz. The speed of sound is 340 m/s. Calculate the difference in frequencies heard by a stationary observer if the ambulance - Physics

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Question

The sound emitted from the siren of an ambulance has a frequency of 1500 Hz. The speed of sound is 340 m/s. Calculate the difference in frequencies heard by a stationary observer if the ambulance initially travels towards and then away from the observer at a speed of 30 m/s.

Numerical
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Solution

Given: vs = 30 m/s, n0 = 1500 Hz, v = 340 m/s

To find: Difference in apparent frequencies (nA − n'A)

Formulae:

  1. When the ambulance moves towards the stationary observer, then `"n"_"A" = "n"_0("v"/("v" - "v"_"s"))`
  2. When the ambulance moves away from the stationary observer, then `"n"'_"A" = "n"_0("v"/("v" + "v"_"s"))`

Calculation: From formula (i),

`"n"_"A" = 1500(3400/(340 − 30))`

∴ nA = 1645 Hz

From formula (ii),

n'A = 1500`(340/(340 + 30))`

∴ n'A = 1378 Hz

Difference between nA and n'

= nA − n'= 1645 − 1378 = 267 Hz

The difference in frequencies heard by the stationary observer is 267 Hz.

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Chapter 8: Sound - Exercises [Page 158]

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Balbharati Physics [English] Standard 11 Maharashtra State Board
Chapter 8 Sound
Exercises | Q 3. (x) | Page 158

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