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Doppler effect Eeeeeee – yowwwwwwwwww A change in frequency or pitch of a sound detected by an observer.

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Presentation on theme: "Doppler effect Eeeeeee – yowwwwwwwwww A change in frequency or pitch of a sound detected by an observer."— Presentation transcript:

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2 Doppler effect Eeeeeee – yowwwwwwwwww A change in frequency or pitch of a sound detected by an observer.

3 Stationary source production of sound waves Velocity of source (v s ) = 0 Both women hear the same sound (pitch)

4 Moving source production of sound waves Velocity vector in direction (v s ) has positive value Hears lower pitch – longer λ hears higher pitch – shorter λ

5 HOW DOES IT WORK??? Let’s start with a review of equations: λ = V (wave) / f units: m (per wave) f = V (wave) / λ units: waves / s T = λ / V (wave) = 1 / f units: s / wave

6 How about a visual here Stationary source: 1 pulse/s (siren)  Moving pulse: 1 pulse/s (siren v =.5m/s)

7 λ = 1m λ’ =.5m λ’ (m/wave) = λ (m/wave) - V siren (m/s) x period (s/wave) λ’ = λ - (V siren)(T) Δ λ = λ – λ’ = λ - λ - V siren (λ/V sound) Δ λ = V siren (λ) / V sound

8 Fun and exciting question for you: If a siren wave frequency is 1600 Hz And the V of the siren is 25 m/s Then what is the frequency of the siren as heard by an observer (as the siren moves toward the observer)?

9 You need the λ and the period to solve λ = V sound / f = 343 / 1600 =.214 m T = 1/f = 1/1600 = 6.25 x 10 -4 s/wave Now calculate λ’ λ’= λ – V siren (T) =.214 m/wave – (25 m/s x 6.25 x 10 -4 s/wave) =.198 m f’ = V sound / λ’ = 343 /.198 = 1732 Hz


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