Doppler effect Eeeeeee – yowwwwwwwwww

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Presentation transcript:

Doppler effect Eeeeeee – yowwwwwwwwww A change in frequency or pitch of a sound detected by an observer. Unless specified – assume that the speed of sound waves in the air is 343 m/s This velocity is a fair estimate under most daily conditions

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

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

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

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

Δ λ = V source (λ) / V sound λ = 1m λ’ = .5m λ’ (in m/wave) = the wavelength that the observer hears as the sound source moves toward them λ’ = λ - V sound source x period λ’ = λ - (V source)(T) And another: Δ λ = V source (λ) / V sound

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

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

One more fun one for you!!! A motorcycle with a wavelength exhaust note of .83m (414 Hz) is moving away from you at 54 m/s. What wavelength do you hear as it moves away? 0.965 m (a frequency of 355 Hz)