This physics worksheet helps students calculate the speed of sound across various mediums like air and steel, accounting for temperature changes.

Let’s go step by step through each question. We’ll use the formula for speed of sound in air:

> v = 331 + 0.6 × T
> where:
> - v = speed of sound (m/s)
> - T = temperature in °C

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Question 1:

> The speed of sound increases with _______ through which it travels. Sound travels faster in ______ than in ______.

✔ Answer:
The speed of sound increases with temperature through which it travels.
Sound travels faster in solids than in gases.

*(Note: Actually, sound travels fastest in solids, then liquids, then gases — because particles are closer together and transfer energy faster.)*

But looking at the context of the worksheet (which focuses on temperature), maybe they want “temperature” for first blank, and then compare media? Let’s check next questions — they’re all about temperature. So perhaps the second part is misphrased? But standard knowledge: sound travels faster in denser media like water or steel than in air.

Wait — let’s read carefully:
“through which it travels” → that refers to medium? Or temperature?

Actually, re-reading:
“The speed of sound increases with _______ through which it travels.” → This is awkwardly worded. Probably meant:
“The speed of sound increases with the temperature of the medium through which it travels.”

Then: “Sound travels faster in ______ than in ______.” → Likely comparing states of matter: solids > liquids > gases.

So:

→ First blank: temperature
→ Second blank: solids
→ Third blank: gases

*(Some might say “liquids” vs “gases”, but solids are even faster. Since this is likely middle/high school level, “solids” and “gases” is acceptable.)*

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Question 2:

> The speed of sound in dry air at 20°C is 343 m/s. At 30°C, the speed of sound would be ______ m/s. For every degree increase in temperature, the speed of sound increases by ______ m/s.

We know from formula:
At 20°C → 331 + 0.6×20 = 331 + 12 = 343 m/s ✔ matches given.

At 30°C → 331 + 0.6×30 = 331 + 18 = 349 m/s

Increase per degree = 0.6 m/s

✔ Answers:
First blank: 349
Second blank: 0.6

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Question 3:

> What is the speed of sound at 15°C?

Use formula:
v = 331 + 0.6 × 15 = 331 + 9 = 340 m/s

✔ Answer: 340

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Question 4:

> What is the speed of sound at -20°C?

v = 331 + 0.6 × (-20) = 331 - 12 = 319 m/s

✔ Answer: 319

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Question 5:

> When does the velocity become negative? Therefore, two important equations can be derived...

This seems incomplete or cut off. Looking at the image description, there’s a box with:

> v = 331 + 0.6T
> and another equation: T = (v - 331)/0.6

Also, it says: “When does the velocity become negative?”

Set v < 0:

331 + 0.6T < 0
→ 0.6T < -331
→ T < -331 / 0.6 ≈ -551.67°C

That’s way below absolute zero (-273.15°C), so physically impossible.

So probably the question is rhetorical or testing understanding — velocity never becomes negative under normal conditions.

But since the worksheet asks to derive two equations, we already have:

1. v = 331 + 0.6T
2. T = (v - 331)/0.6

And for “when does velocity become negative?” — answer: Never under real-world conditions, since temperature can’t drop low enough.

But if forced to give numerical answer: below -551.67°C

However, looking at the boxes in the image, maybe they just want the rearranged formula?

Let me see the structure again.

It says: “Therefore, two important equations can be derived...”

Then shows:

Box 1: v = 331 + 0.6T
Box 2: T = ????

So Box 2 should be: T = (v - 331)/0.6

And for “when does velocity become negative?” — perhaps leave as is or note it's not possible.

But since the student needs to fill blanks, and based on typical worksheets, I think they expect:

For the “when does velocity become negative?” — maybe they want the temperature value? Even though unrealistic.

Alternatively, perhaps it’s a trick question — velocity doesn’t become negative.

But let’s look ahead — question 6 uses these formulas.

I’ll assume for now:

Answer for “when does velocity become negative?” → at temperatures below approximately -552°C (but note: not physically possible)

But since the worksheet may just want the formula derivation, I’ll focus on that.

Actually, rereading: “When does the velocity become negative? Therefore, two important equations can be derived...”

Perhaps it’s implying that when you solve for T, you get the inverse.

I think the key is to write the two equations:

Equation 1: v = 331 + 0.6T
Equation 2: T = (v - 331)/0.6

And for “when does velocity become negative?” — maybe skip or say “never”.

But to match the format, let’s put:

Blank for “velocity becomes negative at”: T < -551.67°C

But that’s messy. Perhaps the worksheet expects only the formulas.

Looking at the image layout, there are three boxes after that sentence.

Probably:

First box: v = 331 + 0.6T
Second box: T = (v - 331)/0.6
Third box: ??? Maybe the condition?

Alternatively, perhaps the third box is for the calculation of when v=0.

Set v=0:

0 = 331 + 0.6T → T = -331/0.6 = -551.67°C

So velocity becomes zero at -551.67°C, negative below that.

So:

“When does the velocity become negative?” → Below -551.67°C

But again, not realistic.

Since this is educational, I’ll provide the mathematically correct answer.

✔ So for Question 5:

- Velocity becomes negative at temperatures below -551.67°C
- Equation 1: v = 331 + 0.6T
- Equation 2: T = (v - 331)/0.6

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Question 6:

> A siren horn emits a sound wave with wavelength 0.8 m. If the speed of sound is 340 m/s, what is the frequency?

Formula:
v = f × λ
where:
v = speed (m/s)
f = frequency (Hz)
λ = wavelength (m)

So:
f = v / λ = 340 / 0.8 = 425 Hz

✔ Answer: 425

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Question 7:

> An echo takes 0.5 seconds to return. How far away is the reflecting surface? Speed of sound is 340 m/s.

Echo means sound goes to the surface and back → total distance = 2d

Time t = 0.5 s
Speed v = 340 m/s

Distance traveled by sound = v × t = 340 × 0.5 = 170 meters

But this is round trip → one-way distance d = 170 / 2 = 85 meters

✔ Answer: 85

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Now, compiling all answers:

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Final Answer:

1. temperature, solids, gases
2. 349, 0.6
3. 340
4. 319
5. Below -551.67°C; v = 331 + 0.6T; T = (v - 331)/0.6
6. 425
7. 85

*(Note: For Q5, if the worksheet expects only the formulas without the temperature value, adjust accordingly. But based on logic, including the temp makes sense.)*

But let’s double-check Q5 wording:
“When does the velocity become negative? Therefore, two important equations can be derived...”

Possibly, the “therefore” implies that from setting v<0, we derive the inverse formula? Not really.

More likely, the two equations are the forward and reverse formulas.

And “when does velocity become negative” is separate.

In many textbooks, they ask: “At what temperature would speed of sound be zero?” → T = -331/0.6 = -551.67°C

So negative below that.

I think it’s fine.

Also, for Q1, some sources say sound travels faster in liquids than gases, but solids are faster still. Since the question says “than in ______”, and no specification, “solids” and “gases” is best.

If the curriculum teaches only liquid vs gas, then “liquids” and “gases”. But generally, solids are included.

Given that, I’ll stick with solids/gases.

Final verification:

Q2: 30°C → 331 + 18 = 349 ✔️
Q3: 15°C → 331 + 9 = 340 ✔️
Q4: -20°C → 331 - 12 = 319 ✔️
Q6: 340 / 0.8 = 425 ✔️
Q7: 340 * 0.5 / 2 = 85 ✔️

All good.

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Final Answer:
1. temperature, solids, gases
2. 349, 0.6
3. 340
4. 319
5. Below -551.67°C; v = 331 + 0.6T; T = (v - 331)/0.6
6. 425
7. 85