Educational worksheet on waves and sound, featuring questions on the Doppler effect, frequency, and speed of sound calculations.

Let’s go through each question one by one. I’ll explain the thinking behind each answer so you understand why it’s correct.

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1. When an automobile moves towards a listener, the sound of its horn seems relatively ___

When something moves toward you, the sound waves get squished together — that means higher pitch (like when a siren comes closer and sounds “higher”). So:
→ high pitched

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2. When the automobile moves away from the listener, its horn seems ___

Now the waves stretch out — lower pitch (like when the siren passes and goes “wheeeeee” as it leaves).
→ low pitched

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3. The changed pitch of the Doppler effect is due to changes in ___

The actual frequency of the source doesn’t change — but because the source or observer is moving, the *observed* frequency changes. That’s what we call “apparent frequency.”
→ c. apparent frequency

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4. Circle the letter of each statement about the Doppler Effect that is true

Let’s check each:

a. It occurs when a wave source moves towards an observer → ✔ True
b. It occurs when a wave source moves away from an observer → ✔ True
c. It occurs when an observer moves towards a wave source → ✔ True (you hear higher pitch if you run toward a speaker)
d. It occurs when an observer moves away from a wave source → ✔ True (you hear lower pitch if you run away)

So all four are true! But since it says “circle the letter,” and doesn’t say “choose one,” we assume multiple can be selected. In many worksheets, they expect you to pick all that apply. So:
→ a, b, c, d

*(Note: If your teacher expects only one, this might be tricky — but scientifically, all are correct.)*

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5. True / False: A moving wave source does not affect the frequency of the wave encountered by the observer.

This is FALSE. That’s literally what the Doppler effect is — motion affects the observed frequency.
→ False

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6. True / False: A higher frequency results from a wave source moving towards an observer.

Yes — waves compress → more waves per second hit your ear → higher frequency.
→ True

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7. Two fire trucks with sirens on speed towards and away from an observer...

Truck A is coming toward you → higher pitch
Truck B is going away → lower pitch

a. Which truck produces a higher than normal siren frequency? → A
b. Which truck produces a lower than normal siren frequency? → B

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8. What is the frequency heard by a person driving at 15 m/s toward a blowing factory whistle (800 Hz) if the speed of sound is 340 m/s?

Use Doppler formula for observer moving TOWARD stationary source:

> f' = f × (v + v₀) / v
Where:
f = original frequency = 800 Hz
v = speed of sound = 340 m/s
v₀ = observer speed = 15 m/s (toward source → positive)

f' = 800 × (340 + 15) / 340
= 800 × 355 / 340
Calculate 355 ÷ 340 ≈ 1.0441
Then 800 × 1.0441 ≈ 835.3 Hz

Round to whole number? Probably fine.
→ 835 Hz

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9. From previous problem, what frequency would he hear after passing the factory?

Now he’s moving AWAY → use minus sign:

f' = f × (v - v₀) / v
= 800 × (340 - 15) / 340
= 800 × 325 / 340
325 ÷ 340 ≈ 0.9559
800 × 0.9559 ≈ 764.7 Hz

→ 765 Hz

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10. Car approaching stationary observer emits 450 Hz. Observer detects 470 Hz. How fast is car moving? Speed of sound = 343 m/s

Source moving TOWARD stationary observer:

Formula:
> f' = f × v / (v - vₛ)
We need to solve for vₛ (source speed)

Given:
f' = 470 Hz
f = 450 Hz
v = 343 m/s

So:
470 = 450 × 343 / (343 - vₛ)

Divide both sides by 450:
470/450 = 343 / (343 - vₛ)
≈ 1.0444 = 343 / (343 - vₛ)

Take reciprocal:
(343 - vₛ)/343 = 1 / 1.0444 ≈ 0.9575

Multiply both sides by 343:
343 - vₛ ≈ 0.9575 × 343 ≈ 328.4

So:
vₛ ≈ 343 - 328.4 = 14.6 m/s

→ 14.6 m/s

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11. Train whistling at 440 Hz, detected at 475 Hz. Speed of sound = 343 m/s. Find train speed.

Again, source moving toward stationary observer:

f' = f × v / (v - vₛ)

475 = 440 × 343 / (343 - vₛ)

Divide both sides by 440:
475/440 ≈ 1.0795 = 343 / (343 - vₛ)

Reciprocal:
(343 - vₛ)/343 = 1 / 1.0795 ≈ 0.9263

Multiply by 343:
343 - vₛ ≈ 0.9263 × 343 ≈ 317.7

vₛ ≈ 343 - 317.7 = 25.3 m/s

→ 25.3 m/s

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12. Determine speed of sound at 45.0°C

Formula:
v = 331 + 0.6 × T (where T is in °C)

v = 331 + 0.6 × 45
= 331 + 27
= 358 m/s

→ 358 m/s

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13. Burglar alarm emits 1200 Hz. Police car drives toward at 40.0 m/s. Temp = 35.0°C. What frequency do police hear?

First, find speed of sound at 35°C:

v = 331 + 0.6 × 35 = 331 + 21 = 352 m/s

Police are moving TOWARD stationary source → use observer-moving-toward formula:

f' = f × (v + v₀) / v
= 1200 × (352 + 40) / 352
= 1200 × 392 / 352

Calculate 392 ÷ 352 ≈ 1.1136
1200 × 1.1136 ≈ 1336.36 Hz

→ 1336 Hz (rounded to nearest whole number)

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Final Answer:
1. high pitched
2. low pitched
3. c. apparent frequency
4. a, b, c, d
5. False
6. True
7. a. A; b. B
8. 835 Hz
9. 765 Hz
10. 14.6 m/s
11. 25.3 m/s
12. 358 m/s
13. 1336 Hz