SME
Grade 11 Science · Problem type
The Doppler effect
Use this for “A 700 Hz siren passes you at 25 m/s. What pitch do you hear as it comes and as it goes?”
Diagram
Ahead: f′ = f v/(v − vₛ) = 700 × 343/(343 − 25) = 755 Hz
Behind: f′ = f v/(v + vₛ) = 700 × 343/(343 + 25) = 652 Hz
Fronts bunch up ahead (shorter wavelength, higher pitch) and spread out behind (lower pitch).
This is an example. Type your own values to start a new problem.
f′₁ (Hz)
=
f (Hz)
×
v (m/s)
v (m/s)
−
vₛ (m/s)
λ
Wavelength
calculated
f′₂
Frequency heard behind
calculated · Hz
Formulas
λ
=
v
f
0.49
=
343
700
f
′₁
=
f
×
v
/(
v
−
vₛ)
755 = 700 × 343/(343 − 25)
f
′₂
=
f
×
v
/(
v
+
vₛ)
652.4 = 700 × 343/(343 + 25)
Assumptions
•
The source sends out one wavefront each period, from wherever it is then; the air carries each out at v.
•
Ahead the fronts bunch up: a higher frequency. Behind they spread out: a lower one.
•
The listener stands still (one moving toward a still source hears f(v + vL)/v); the source moves slower than sound.
Step-by-step
WE KNOW
vₛ = 25 m/s, v = 343 m/s, f = 700 Hz
FIND
wavelength (λ), frequency heard ahead (f′₁), frequency heard behind (f′₂)
Step 1 · Find wavelength (λ)
Use λ = v/f
At
rest
the
waves
are
v
f
apart.
λ
=
343
700
λ
=
0.49
m
Step 2 · Find frequency heard ahead (f′₁)
Use f′₁ = f × v/(v − vₛ)
Ahead
the
fronts
are
only
(
v
−
vₛ)/
f
apart:
a
higher
pitch.
f
′₁
=
700
×
343/(343
−
25)
f
′₁
=
700
×
343
318
f
′₁
=
240,100
318
f
′₁
=
755
Hz
Step 3 · Find frequency heard behind (f′₂)
Use f′₂ = f × v/(v + vₛ)
Behind
the
fronts
are
(
v
+
vₛ)/
f
apart:
a
lower
pitch.
f
′₂
=
700
×
343/(343
+
25)
f
′₂
=
700
×
343
368
f
′₂
=
240,100
368
f
′₂
=
652.4
Hz
Check
Put the numbers back in. Both sides should match.
0.49
=
343
700
✓
755 ≈ 700 × 343/(343 − 25) ✓
652.4 ≈ 700 × 343/(343 + 25) ✓
Topic check
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