Help me out with this physics problem

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SUMMARY

The discussion focuses on calculating the spacing between uprights of an iron railing based on a sound frequency of 640 Hz and the speed of sound in air, which is 330 m/s. The formula used is λ = v / f, where λ represents wavelength, v is velocity, and f is frequency. The problem involves understanding that the sound reflects off the bars, creating a total return frequency. To find the spacing, users must consider the round trip time of the sound waves.

PREREQUISITES
  • Understanding of wave properties, specifically wavelength and frequency
  • Familiarity with the speed of sound in air
  • Basic algebra for manipulating equations
  • Knowledge of sound reflection principles
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  • Calculate the wavelength using the formula λ = v / f with v = 330 m/s and f = 640 Hz
  • Explore the concept of sound reflection and its effects on perceived frequency
  • Research the speed of sound variations under different atmospheric conditions
  • Learn about wave interference and its implications in sound propagation
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Students studying physics, educators teaching sound wave properties, and anyone interested in acoustics and wave behavior.

woolly
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I have been scratching my head all night about this problem...

A man standing close to an iron railing consisting of evenly-spaced uprights makes a sharp sound and hears a note of frequency 640hz. calculate the spacing between the uprights. (speed of sound in air = 330m/s)


perhaps one of you can help me out with this problem...


THANKS!
 
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It is not clear to me that there is any relationship between the bars and the sound. But

[tex]\lambda = \frac v f[/tex]
where
[tex]\lambda= {Wavelength}[/tex]
[tex]v= {velocity}[/tex]
[tex]f = {frequency}[/tex]

Is what you need.
 
Integral said:
It is not clear to me that there is any relationship between the bars and the sound. But

[tex]\lambda = \frac v f[/tex]
where
[tex]\lambda= {Wavelength}[/tex]
[tex]v= {velocity}[/tex]
[tex]f = {frequency}[/tex]

Is what you need.

I get the problem. The sharp sound is a single discrete narrow pulse, and it echos of the bars, one after another, so that it's total return to the observer has the given frequency.

To solve this problem, there is one more bit of information you'll need to know, and that's the speed of sound, which if I recall is about 770 mph at STP (convert that to m/s). Use that for v to get the bar spacing. Don't forget it's round trip time you'll be working with.

I'm not 100% certain this is what the problem intends, so maybe you should desribe your assumptions in the answer to the homework problem. This way you can get credit even if the instructor intended something else.
 

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