a tuning fork of frequency 200 hertz will resonate if a sound wave incident on it has a frequency of?
455977
200 Hertz or any integer multiple thereof.\nSo, 400 Hertz, 600 Hz, etc.. would cause it to resonate.
289044
The frequency range is generally accepted to be 20 Hz to 20,000 Hz. With the speed of sound = 334 m/s, 20 Hz has a wavelength of 16.7 m, 20,000 Hz a wavelength of 16.7 mm. (Ubergeek, the answerer in your link slipped a decimal place.)
683354
b. 93,300,000 Hz\n\nFM stations measure their frequencies in Megahertz (MHz)
325938
Well, I SINCERELY hope you are NOT successful!!\n\nEvery violin in the world is quite unique in resonance.\nReasons for this include the density of the woods used, the age of the violin, construction methods and humidity history over the life of the instrument.\n\nYou will find, however, that the frequencies of one octave around middle-c will have the HIGHEST resonance effect.\nA VERY strong SQUAREWAVE ( +/1 80db ) of around 260Hz should cause structural damage to the violin AS WELL AS YOUR EARS!! ( Squarewave because of the rich harmonics )
566130
It just is.  Many countries use 50 hertz.  It is one of the standard frequencies.   The US uses 60.
152237
Try this: http://www.piezo.com/index.html
1106409
A subwoofer is a type of loudspeaker dedicated to the reproduction of bass frequencies, typically from about 20 Hz to about 200 Hz. It is difficult for small loudspeakers to reproduce frequencies below 40 Hz, especially above 100 dB, and so it is often advantageous to use a loudspeaker intended specifically for this task.
693965
If it is open on both ends, the harmonic frequencies are:\n\nf=nv/2L\n\nwhere n is the harmonic number (1st, 2nd, etc), v is the speed of sound in air (~343 m/s), and L is the length of the pipe.
164646
You mean a 'frequency of 5.10^12Hz'\n\nCan you work it out yourself?:\n\nRemember that the speed (v) of the wave is 3×10^8 m/s,\nand that the wavelength is given by v/f\n\n[Edit: Darn, Kirchwey - don't do Glow's homework for her! ;-) ]
1112020
okay, i found it.\n\n" While a tuning fork used in music is tuned to 440 hertz (the “A” note), seismic instrumentation and analysis determined the Earth’s free oscillation to be composed of pitches between 2 to 7 millihertz, fifty notes together packed within less than two octaves, ringing about 16 octaves below mankind’s musical “Middle C.”"\n\nso, 16 octaves BELOW mankind's musical middle c.\n\ni guess God has a really deep voice.\nvery cool.
1047677
no, but the vibrations, since they're so fine, can completely screw up the nerves inside and give you one hell of a toothache. i've done it with things other than a tuning fork and it was really bad. but if you actually hit your tooth with it, it probably can. though it would have to be really, really hard
143763
a signal emmited in the band of the 2 GHz.
647307
Try the sonometer or the resonance experiment
945463
The resonant frequencies for a pipe closed at one end is given by\n\nf = n*V/(4*L)\n\nwhere V is the speed of sound, L is the length of the pipe, and n is an odd integer.\n\nFor your case\n\nf1 = n1*V/(4*L)\nf2 = n2*V/(4*L) = (n1+2)*V/(4*L)\n\nf2 - f1 = 2*V/(4*L) = V/(2*L)\n\nSolving for L\n\nL = V/(2*(f2-f1))\n\nL = (340 m/s)/(2*100 Hz) = 340/200 m = 1.7 m\n\nwhere f1 and f2 were chosen to be 550 Hz and 650 Hz, respectively.
663539
Most solids tend to contract slightly when they get cold.  They also get more brittle.  My own experience as a cellist is that when my instrument is cold, the strings get tighter and thus the pitch gets higher.  Although I've never tried it with a tuning fork, I suspect the same thing would happen.  If the cold causes it to be smaller and less malleable, it would probably change the natural resonant frequency of the tuning fork as well, moving it up a notch.
82505
First, you need to find the wavelength of the frequency.\nSpeed=Wavelegth*frequency\nWavelength= 342 m/s / 440 1/s =.78 meters\n\nThe wave traveling into the tube will hit its first resonance frequency one quarter of the way along the wave.  This is the first spot where that wave travels into the tube at such an angle that when it bounces off the water, it can escape the tube again; that is what creates the "resonance" or the louder vibrating sound.  So the first resonance would be @ .78m/4=.19m ; the second @ .19*2=.39m ; and the third at .19*3=.58 meters.
168920
This is ultrasonic, not electromagnetic, so C is irrelevant.  The equation is:\nF=V/L where V=sound velocity in water (~1475 m/s), F=frequency in Hz, L=wavelength (0.0003 m)\nResult is you need 4.9 MHz or higher.