Speed of Sound Calculation in Air Physics

Physicshelp Canada
8 Dec 201202:43

Summary

TLDRThis educational script explains how the speed of sound varies with temperature, using a real-world scenario. On a cold day at 4°C, the narrator calculates the distance to a mountain by measuring the time it takes for their voice to echo back, 2.8 seconds. Using the formula V = 331 + 0.59 * T°C, they determine the sound speed as 333 m/s. By applying kinematic equations, they find the mountain is 466 meters away, illustrating the practical application of physics in everyday life.

Takeaways

  • 🌡️ The speed of sound varies with temperature; it travels faster in warmer air.
  • ❄️ On a cold day, sound travels slower, while on a warm day, it travels faster.
  • 📐 The formula to calculate the speed of sound in air is V = 331 + 0.59 * temperature in degrees Celsius.
  • 🔊 To determine the distance to an object, like a mountain, one can use the formula D = V * t, where V is the speed of sound and t is the time.
  • ⏱️ The time it takes for sound to travel to an object and back is twice the time it takes for sound to travel one way.
  • 🏔️ In the example, the temperature was 4°C, which made the speed of sound approximately 333 m/s.
  • 📏 The total distance the sound waves traveled to the mountain and back was calculated to be 933 meters.
  • 🔄 The actual one-way distance to the mountain was half of the total distance, which is 466 meters.
  • 🧮 The problem demonstrates the application of kinematic equations and the speed of sound formula to solve real-world scenarios.
  • 🤔 The script encourages critical thinking by combining physics concepts to solve a practical problem involving sound propagation.

Q & A

  • How does temperature affect the speed of sound in air?

    -The speed of sound in air increases as the temperature rises. On a cold day, sound travels slower, while on a warm day, it travels faster.

  • What is the formula used to calculate the speed of sound in air?

    -The formula used to calculate the speed of sound in air is V = 331 + 0.59 * (temperature in degrees Celsius).

  • If it's 4°C outside, how fast does sound travel according to the formula?

    -At 4°C, the speed of sound is approximately 333 m/s, calculated as 331 + 0.59 * 4.

  • What is the kinematic equation used to determine the distance sound waves travel?

    -The kinematic equation used is V = D/t, where V is the speed of sound, D is the distance, and t is the time.

  • How long did it take for the sound to travel to the mountain and back in the example?

    -In the example, it took 2.8 seconds for the sound to travel to the mountain and back.

  • What is the total distance the sound waves traveled in the example?

    -The sound waves traveled a total distance of 933 meters, calculated by multiplying the speed (333 m/s) by the time (2.8 seconds).

  • Why is the total distance calculated in the example twice the actual distance to the mountain?

    -The total distance is twice the actual distance to the mountain because the sound waves traveled to the mountain and then reflected back.

  • What is the actual distance to the mountain in the example?

    -The actual distance to the mountain is 466 meters, which is half of the total distance the sound waves traveled (933 meters).

  • How does the example demonstrate the use of both the speed of sound equation and the kinematic equation?

    -The example demonstrates the use of both equations by first calculating the speed of sound at a given temperature and then using that speed to determine the distance using the kinematic equation.

  • What is the significance of using both equations in the problem-solving process?

    -Using both equations allows for a comprehensive approach to solving problems involving the speed and distance of sound waves, particularly when environmental conditions like temperature are variable.

Outlines

00:00

🌡️ Effect of Temperature on Sound Speed

This paragraph explains how the speed of sound waves in air is influenced by temperature. On cold days, sound travels slower, while on warm days, it travels faster. The speed of sound in air is given by the formula V = 331 + 0.59 * temperature in degrees Celsius. The script uses a practical example where the speaker calculates the distance to a mountain by measuring the time it takes for sound to travel to the mountain and back, using the speed of sound formula adjusted for the temperature of 4°C. The calculation results in the mountain being 466 meters away.

Mindmap

Keywords

💡Sound Waves

Sound waves are mechanical waves that propagate through a medium, such as air, by the vibration of particles. In the video, the concept is central as it explains how these waves travel faster in warmer air and slower in colder air, affecting the speed at which sound travels. The example used is a scenario where the speaker yells on a cold day, and the sound waves reflect off a mountain, illustrating the practical application of understanding sound wave propagation.

💡Speed of Sound

The speed of sound refers to how fast sound waves travel through a medium. It is a fundamental concept in the video, with a specific formula provided to calculate it in air: V = 331 + 0.59 * temperature in degrees Celsius. This formula is used to determine the speed of sound on a cold day, which is crucial for calculating the distance to the mountain in the example.

💡Temperature

Temperature plays a significant role in the speed at which sound waves travel. The video script mentions that sound waves travel faster as the air gets warmer, which is why temperature is a key factor in the speed of sound formula. The example given is a cold day at 4 degrees Celsius, where the temperature affects the speed calculation for the sound waves.

💡Kinematics Equation

The kinematics equation is a fundamental principle in physics used to describe the motion of objects. In the video, the equation V = D/t (velocity equals distance divided by time) is rearranged to solve for distance, which is essential for calculating the distance to the mountain based on the time it takes for the sound to travel and return.

💡Uniform Motion

Uniform motion is a type of motion where an object moves at a constant speed in a straight line. The video uses the concept of uniform motion to apply the kinematics equation to sound waves, assuming they travel at a constant speed, which is a valid assumption for the problem presented.

💡Reflection

Reflection in the context of the video refers to the bouncing back of sound waves when they encounter a surface, such as a mountain. The script describes a scenario where the speaker's voice is reflected off a mountain, and the time it takes for the sound to return is used to calculate the distance to the mountain.

💡Distance

Distance is a measure of the space between two points. In the video, the speaker calculates the distance to a mountain by timing how long it takes for sound waves to travel to the mountain and back. The concept is integral to understanding the problem-solving process in the script.

💡Time

Time is a measure of the duration between two points. In the video, the time it takes for the sound waves to travel to the mountain and back (2.8 seconds) is a key piece of information used to calculate the distance using the kinematics equation.

💡Velocity

Velocity is defined as the rate of change of an object's position with respect to time. In the video, velocity is calculated using the speed of sound formula and is used to determine the distance the sound waves travel before returning to the speaker.

💡Round Off

Round off refers to the process of approximating a number to a certain level of precision. In the video, the speaker rounds off the calculated speed of sound to 333 m/s for simplicity, which is a common practice in practical calculations to make them more manageable.

💡Calculation

Calculation is the process of computing or estimating numbers. The video involves several calculations, including the speed of sound and the distance to the mountain, which are essential for solving the problem presented. The calculations are based on the principles of physics and the given data.

Highlights

Sound waves travel faster in warmer air.

On a cold day, sound travels slowly compared to a warm day.

The speed of sound is not constant and depends on temperature.

The formula to calculate the speed of sound in air is given by V = 331 + 0.59 * temperature in °C.

A practical problem is presented involving sound waves reflecting off a mountain.

The time it takes for the sound to travel to the mountain and back is 2.8 seconds.

The kinematics equation V = D/t is used to calculate the distance the sound waves travel.

The speed of sound at 4°C is calculated to be approximately 333 m/s.

The total distance traveled by the sound waves is 933 m, considering the round trip.

The actual distance to the mountain is half of the total distance, which is 466 m.

The problem demonstrates the application of both kinematics and speed of sound equations.

The solution involves rearranging the kinematics equation to solve for distance.

The importance of considering the round trip of sound waves in the calculation is highlighted.

The problem showcases a real-world application of physics in determining distances using sound.

The solution requires understanding the relationship between temperature and the speed of sound.

The problem illustrates the concept of sound wave reflection and its use in measuring distances.

The calculation involves rounding off the speed of sound for simplicity.

The final answer is obtained by dividing the total distance by two to get the one-way distance to the mountain.

Transcripts

play00:01

when sound waves travel through air they

play00:04

travel faster as the air gets warmer so

play00:07

on a cold day sound travels very slowly

play00:09

on a very warm day it travels quickly so

play00:11

there's not a constant speed of sound

play00:14

the formula that tells us how fast sound

play00:16

waves are traveling in air is V the

play00:19

speed is equal to 331 + 0.59 * the

play00:23

temperature in degrees C so let's try a

play00:26

very simple problem using that formula

play00:29

I'm standing over here on a very cold

play00:32

day it's 4° C out and I yell and I hear

play00:36

sound waves reflected off a mountain

play00:38

nearby and come back to me 2.8 seconds

play00:40

later and I want to know how far is that

play00:43

mountain from me okay well if I wonder

play00:45

how far it is I'm going to use V is D /t

play00:49

our kinematics equation for uniform

play00:51

motion because the sound waves travel at

play00:53

a constant speed so if I rearrange that

play00:56

I've got the distance that the waves

play00:58

travel is how fast the w waves were

play01:00

traveling times how long they were

play01:01

traveling for I know this this was 2.8

play01:05

seconds that's how long the waves were

play01:07

traveling but how fast were they

play01:08

traveling well that's where this formula

play01:10

might come in I need to know how fast

play01:12

the waves are

play01:13

traveling V =

play01:16

331 +

play01:19

0.59 * the temperature in degrees

play01:22

C 4 so the waves were actually traveling

play01:27

at uh 300

play01:30

33

play01:32

m/s I rounded it off a little bit so the

play01:36

temperature being 4° made the waves

play01:38

travel a little faster so they're

play01:39

traveling at 333

play01:41

m/s now I can plug into this formula I

play01:45

can say the distance the wave travels

play01:47

were was its velocity how fast 333 times

play01:51

how long were they traveling for 2.8

play01:54

seconds so the total is uh

play01:58

93 3 m that's how far the wave traveled

play02:04

but keep in mind the distance it

play02:05

traveled was to the mountain and back

play02:08

and similar to one of the wave questions

play02:10

we did earlier that's twice the distance

play02:13

that I'm looking for so I have to take

play02:15

this

play02:16

distance uh the three um the distance

play02:19

the

play02:20

933 divide by two because that's the

play02:23

trip there and back and I get uh

play02:28

466 m

play02:30

so the distance to the to the mountain

play02:32

is 466 M so this problem used our

play02:36

kinematics equation and our speed of

play02:38

sound equation just to mix it up and

play02:40

make it a little more

play02:41

complex

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Related Tags
Speed of SoundPhysics CalculationTemperature EffectSound WavesDistance MeasurementKinematicsAcousticsOutdoor ExperimentScience EducationPhysics Challenge