A piano emits frequencies that range from a low of about 28 Hz to a high of about 4200 Hz.  Find the range of wavelengths in air attained by this instrument when the speed of sound in air is 340 m/s.
127304
According to the data given\nv= 340 m/s\nf = 28 waves.sec,\nv=fλ \n340=28λ\nAnd Hence\nλ=12.14 m to .08 m
520255
find a piano teacher. they can determine your range, and what keys you sing in.\nThe teacher may know of a vocal coach to help you, or a choir director from a church, also.
525974
Don't give up too easily on the sound wave thing.  Graphic equalization.  Sound is measured in hertz (Hz) & Kilo Hertz (KHz), Kilo being the metric measure for 100.  Start with the low tones - long wavelengths, low number of hertz.  Move up the tone scale. As the tones get higher pitched the smaller the wave lengths - increased number of hertz. Super highs, like cymbals are in the kilohertz range and so on.  It's an everyday thing that everyone can relate to .  When you hear a siren go by the tone is lower as it approches and higher, faster, as it passes you and back to lower tones as the object receeds.  Good luck.
452858
The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave. \n\nThe equation that relates wavelength and frequency for electromagnetic waves is: λν=c where λ is the wavelength, ν is the frequency and c is the speed of light.
637306
Sound travel in air with a speed of 331.4 m/s. For sound to travel from the student to the wall and back it must travel 165*2=330 m.  For sound to travel 330 m , it will take approximately 330/331.4=..9958s.
213606
Electromagnetic radiation in the 1 to 1000 m wavelength range.
512158
KNOWN:\nf = 3.2 x 10^-2 Hz\nc (ie speed of light) = 3.00 * 10 ^8 m/s\n\nFIND L (AKA wavelength)\nL m = c m/s  /  f Hz\n= 3.00 * 10 ^8 m/s  /  3.2 x 10^-2 Hz\n= 9 375 000 kilometers\n\nAre you sure its 10^-2 Hz? That seems unlikely
186125
The speed of light in a vacuum is exactly 299,792,458 metres per second (or 1,079,252,848.8  km·h-1, which is approximately 186,282.397 miles per second, or 670,616,629.4 miles per hour).
235144
315mHz \n\nLove,\nKenneth Horowitz
369439
You have stated that the tuning fork B is loaded and has a lesser frequency than that of A.\n\nLet the frequency of B be N and you know that N is lesser than 256 Hz.\n\nBeat frequency is the difference between the two frequencies.\n\nTherefore 256 - N = 10 ( given).\n\nOr N = 246 hz.\n\nFor your second  question:\n\nVelocity = frequency X wavelength.\n\nIn a closed tube the first resonant length is obtained when the length is one fourth of the wavelength.\n\n Therefore L = 4 x l.\n\nOr V = 4 N l where N is the frequency and l is first resonating length.\n\n\n344 =  4 x 440 x l.\n\nl = 344 / 1760 = 0.195 m  = 19.5 centimeter.
413776
length = 3.5m (appr)\n\nn=6\n\n(hint: find GCF of frequencies)
455171
The frequency in the number of times the wave vibrates in one second. \nIn your question, you probably have light.  The speed of light is 300,000km/s.  If you multiply the frequency and the wavelength, you get the speed.  That is v=lf\nThat 595nm is actually 595x10e-9m or 0,000000595m\nThe speed of light is 3x10e8m/s\nSo the frequency is\nf=v/l\nf=300000000/0.000000595\nf=504201680672268.91\nf=5.1e14 hertz
433143
If the period is 2.60 seconds, the frequency is 2Pi/2.60 = 2.42 radians/sec. The natural frequency is sqrt(k/m) (see the reference).\n\n2.42 = sqrt(k/m) = sqrt(k/7)\n\n5.84 = k/7\n\nSo the spring constant k = 40.88
188316
the wavelength is half...\n\nvavelength and frequency are inversely proportional
527663
get old fashion tuning forks, and tune it by your ear. That is how they were tuned for hundreds of years, and regardless, it will sound right to you!\n\nIn 1975 my friend introduced me to tunning pianos. I had NO clue, but could match the strings to the sound of the tunning fork.
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If you can measure the *frequency*, you know the wavelength.\n(As the wavelength is just the frequency divided by the speed of sound.)  I found a calculator that does this for you:\nhttp://www.sengpielaudio.com/calculator-wavelength.htm\n\nIdeally you would get an oscilloscope to measure the frequency.\n\nIf you don't have access to one, if you have a guitar tuner, you could measure the frequency (the tuner will tell you the pitch, say A#, and you can look up the frequency of that pitch in a music book).\n\nOr if you have a bunch of tuning forks or a pitch pipe, you could get an approximation.\n\nIf you have an ocilloscope, don't forget to get the precise speed of sound at your elevation (sound travels a bit faster at sea level, than if you live in, say, Denver).  (Note that the calculator link above gives you a place to enter the speed of sound where you live.)  If you're using a guitar tuner or tuning forks, they are only approximations anyway, so the exact speed of sound for your elevation won't make a difference.  \n\nP.S.  I found a reference to a sound analyzer you can get for your computer!  The software is free, but it needs a soundcard in your computer (which you probably have).  It uses the information from a microphone in your computer to give you the frequency and wavelength of the musical note.  It's kinda cheating if this is for a science-fair project ... but maybe this is what you want.\nhttp://www.relisoft.com/Freeware/freq.html
466465
wavelength*frequency=c\n\n3.04667 m