Calculating Distance Traveled by a Car's Sound Emission

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SUMMARY

The discussion centers on calculating the distance a car travels during one period of sound emitted from its horn. A car moving at 30 m/s emits a 220 Hz sound, resulting in a sound period of 4.54 ms. During this time, the sound travels 1.55722 meters, while the car moves 0.1362 meters. The calculations confirm that the apparent wavelength varies based on the observer's position relative to the car.

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  • Understanding of sound wave properties, specifically frequency and wavelength
  • Knowledge of basic physics concepts such as speed, distance, and time
  • Familiarity with the formula for calculating wavelength: λ = v/f
  • Ability to perform unit conversions, particularly between milliseconds and seconds
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Eng67
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I am having a problem with the following question.

A car is moving at 30 m/s. The cars horn is emmitting a 220 hz sound. The speed of sound is 343 m/s. How far does the the car move in one period of the sound emitted from the horn.

I am having a problem determining the distance formula.
The period of the 220 hz sound is 1/220hz = 4.54 ms

so in 4.54 ms the sound has traveled 1.55722 meters (343 * .00454)

Now the distance the car travels would be (4.54 ms * 30 m/s) = .1362 meters

Is this the correct way to go about this problem?
 
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Eng67 said:
I am having a problem with the following question.

A car is moving at 30 m/s. The cars horn is emmitting a 220 hz sound. The speed of sound is 343 m/s. How far does the the car move in one period of the sound emitted from the horn.

I am having a problem determining the distance formula.
The period of the 220 hz sound is 1/220hz = 4.54 ms

so in 4.54 ms the sound has traveled 1.55722 meters (343 * .00454)

Now the distance the car travels would be (4.54 ms * 30 m/s) = .1362 meters

Is this the correct way to go about this problem?
Yes. 1.557 meters is the wavelength of the sound: [itex]\lambda = v/f[/itex]. The apparent wavelength depends on where the observer is. If the observer is behind the car, the wavelength appears to be 1.557+.1362. If the observer is in front, it is .1557 - .1362.

AM
 
Thanks!

I thought I was correct in my calculations.
 

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