Physics review sheet featuring problems on calculating the speed of sound and echo times based on air temperature.

Let’s go step by step through each question on the worksheet. I’ll solve them one at a time, check my work, and then give you the final answers clearly.

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Question 1:

> The speed of sound in air depends on temperature. At 0°C, the speed is about 331 m/s. For every degree Celsius above 0°C, the speed increases by about 0.6 m/s.

We’re told:
- Base speed at 0°C = 331 m/s
- Increase per °C = 0.6 m/s

So for any temperature T (in °C), the formula is:

Speed = 331 + 0.6 × T

That’s our key formula for all these problems.

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Question 2:

> What is the speed of sound at 20°C?

Use the formula:

Speed = 331 + 0.6 × 20
= 331 + 12
= 343 m/s

✔ Check: 0.6 × 20 = 12 → 331 + 12 = 343 → correct.

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Question 3:

> What is the speed of sound at -10°C?

Speed = 331 + 0.6 × (-10)
= 331 - 6
= 325 m/s

✔ Check: 0.6 × 10 = 6 → subtract because it’s negative → 331 - 6 = 325 → correct.

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Question 4:

> Other than the velocity/temperature equation, there are two important equations involving the speed of sound.

They show:

v = fλ
and
λ = v / f

Then they ask:

> If the tuning fork has a frequency of 512 Hz, what is the wavelength if the speed of sound is 343 m/s?

Use: λ = v / f

λ = 343 / 512

Let’s calculate that:

343 ÷ 512 ≈ ?

Do division:

512 × 0.6 = 307.2
343 - 307.2 = 35.8
Now 35.8 / 512 ≈ 0.07

So total ≈ 0.6 + 0.07 = 0.67

More precisely:

343 ÷ 512 = ?

Use calculator-style:

343 ÷ 512 = 0.669921875...

Round to 3 decimal places: 0.670 meters

Or we can write as 0.67 m if rounding to two decimals.

But let’s keep more precision for now — maybe the problem expects exact fraction or rounded value.

Actually, 343/512 is already simplified? Let’s see:

GCD of 343 and 512?

343 = 7³
512 = 2⁹ → no common factors → so fraction is 343/512 m

But usually we want decimal.

In physics class, often round to 2 or 3 sig figs.

Given 343 and 512 both have 3 sig figs → answer should have 3.

343 ÷ 512 = 0.6699... → rounds to 0.670 m

✔ So wavelength = 0.670 meters

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Question 5:

> How many seconds will it take an echo to reach your ears if you yell into a mountain 50 m away on a day when the air temperature is 20°C?

First, find speed of sound at 20°C → we already did this in Q2: 343 m/s

You yell toward a mountain 50 m away → sound travels TO the mountain (50 m) and BACK to you (another 50 m) → total distance = 100 m

Time = distance / speed

t = 100 / 343

Calculate:

100 ÷ 343 ≈ ?

343 × 0.29 = 343 × 0.3 = 102.9 → too big
343 × 0.29 = 343 × 0.3 - 343 × 0.01 = 102.9 - 3.43 = 99.47
Close to 100

100 - 99.47 = 0.53
So add 0.53 / 343 ≈ 0.0015 → total ≈ 0.2915 s

Better to compute directly:

100 ÷ 343 ≈ 0.291545...

Round to 3 sig figs: 0.292 seconds

✔ Final answer: 0.292 s

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Now, compiling all final answers:

──────────────────────────────────────

Final Answer:
2. 343 m/s
3. 325 m/s
4. 0.670 m
5. 0.292 s