cicadas produce a buzzing sound that has a wavelength in air of 2.69 m.  if the speed of the sound in air is 346 m/s, what is the frequency of the sound produced by a cicada? what is the period?\n\ni need to know how i do this so what steps do i even use? i don't get it.
214622
Frequency is in Hertz which is cycles/second, and the period is the length of each cycle in seconds, so they are related.\n\nThe wavelength is 2.69 m/cycle and speed of sound is 346m/s.  So divide the second by the first to get cycles per second -- look at the units and see what you have to do to get cycles on top and seconds on the bottom, and cancel out meters.\n\n346/2.69 = 128.6 Hz\n\nThe period is the inverse of this, seconds/cycle \n\n1/128.6 = 0.00777 seconds
210718
All light has a velocity of "c."  c is ~ 3*10^8m/s.  With this knowledge you can say that frequency is the inverse of time/distance.  Thus, you can use the formula C(meters/sec) / Freq.(cycles/sec) = Wavelength(meter/cycle) to solve the question.\n\nDo your own work from there.
379623
You might think 200 Hz at first since 900-700 = 200. But the question is how do you get the 700 resonance from a 200 Hz fundamental frequency?  A musical instrument wouldn't do this -if it had 200 Hz as a fundamental frequency, then it would have 400, 600, 800, 1000 ... as resonances, not 700 and 900. \n\nOne answer might be if the object is a closed cylinder. The sound propagates thought the air column and is reflected by the closed end. Thus only odd harmonics are generated (see the link for a drawing to help understand this). So if you have closed column with a fundamental frequency of 100 Hz, then you would see 100, 300, 500, 700, 900 ... Hz resonances - so you can get the 700 and 900 Hz resonances that you describe with nothing in between!!
278559
FORMULA: speed = frequency * wavelenght (sorry cant get symbols to work on yahoo)\n\n(1) 0.2= 1/0.5 * w   ...(as f=1/T)\n=> 0.2=2w\n=> w=0.1\n\n(2) s= 25*(10^8) * 0.12    ...(12cm= 0.12m)\n=> s= 3*10^8\n=> s= 300000000 m/s\n\n(3) 2= f * 0.5\n=> f= 2/0.5\n=> f= 8\n\n8 waves strike dock each second.\n\npretty easy questions.
446739
The phase difference at the point under consideration is an odd multiple of pi, as the vibrations annul each other.\n\nThe phase difference is given by the equation 2pi (R1 - R2)/ L where L is the wavelength of the wave, R1 and R2 are the distances of the point from the sources of the sound.\n\nC = N L where C is the velocity of the wave, N is the frequency and L is the wave length.\n\nTherefore, 2pi (R1 - R2) N = (2 k +1) pi C, where k = 0. 1, 2, 3, etc.\n\nTherefore 2 (R1 - R2) 500 = (2 k +1)350.\n\n(R1 - R2) = (2k +1) x 0.35 m\n\nThe distances will be 5 plus or minus (2k +1) x 0.35 m where k = 0, 1, 2, 3, etc.\n\n\nIf k = 0, the distance will be 5.35 or 4.65 m
380789
L+ e = C/4N\n\nL is the resonance length.\n\ne is the end correction \n\nC is the speed of sound in air\n\nN is the frequency of the fork.\n\nThe above equation can be written as \n\n4NL + 4Ne = C.\n\n\nFor first fork, 4N = 1024.\n\n\n1024 x 0.32 + 1024 e = C --------------- (1)\n\nFor the next fork, 4N = 1536\n\n1536 x 0.208 + 1536 e = C----------------- (2).\n\nEquating (2) and (1) \n\n1536 x 0.208 + 1536 e = 1024 x 0.32 + 1024 e \n\n512 e = 327.68 - 319.49 = 8.19 \n\ne = 0.016 m = 1.6 cm.
131184
For this you should know that sound requires a medium to travel and is facilitated by wavwfront formation. These wavefronts (or disturbances) are in form of varying pressure inside the medium (in our case air). The disturbances (or waves) in form of high and low pressure zone travels at what is called the speed of sound in the air originating from source (the Jet engine in this case).\n\nNow when some jetplane approaches the speed of sound it is acually trying to travel with the high and low pressure zones created by himself. Along the length of the aircaraft itself there is a lot variation in air pressure. This disturbs whole of the aerodynamics of the plane and it becomes very difficult to overcome the resulting drag.\n\nThis is what is called a sound barrier
433168
Listening to music always helps!\nw=w’*t, w=500rpm=500/60 rps=2pi*25/3 rad/s, t=2.6s, then \nw’=w/t=2*pi*25/3/2.6=20.14 rad/s^2;\nu=w’*t^2/2=w*t/2=25/3*2.6/2=10.83=~10 full revs
221888
If one wave equals one cycle, then the answer is:\n\nA. 6 Hz
460249
12.25cycles/20sec = 0.6125 cycles/sec, = 0.6125Hz, which is the frequency.  The period is the reciprocal of the frequency, which equals approx. 1.6327 seconds.\n\nI didn't count that first half cycle because it looks longer than the rest.  I started counting at the first positive zero-crossing at time 2.6, and counted 12.25 cycles from there to time 22.6\n\nThis is coming from someone who has made his living for the last few decades looking at and interpreting thousands of such displays.
90206
Thunder is the sound of the shockwave caused when lightning instantly heats the air around it to up to 30 000 °C (54 000 °F). That super-heated air expands rapidly, then contracts as it cools. The rapid expansion/contraction generates sound waves, making the sound that is called "thunder."\n\nSince sound and light travel at different speeds through the atmosphere, one can time the interval between them to roughly estimate how far away the bolt of lightning is. The speed of sound in air is approximately 332 m/s (742.6 mph), while the speed of light is so fast that the lightning is seen only a few microseconds after the event, so the lightning is approximately one kilometer distant for every 3 second interval (one mile for every 5 seconds).
107346
Never mind. No question is a stupid question as long as u learn from it. Here's my explanation. Where does sound come from? Sound is actually produced by vibration. As long as it has vibration, it produces sound, no matter how loud or how soft. Even molecules in a solid produces sound because it vibrates, but we can't hear the sound because it is out of our hearing range. Well, as you asked, when a tree falls, it actually vibrates, hence, sound is actuallly produced although we can't hear it. For your second question, IF, lets say you are in the woods and happen to see the tree fall, and you can hear the sound of the fall of the tree. Your question was, how on earth you heard that sound, right? Here's the explanation. When the tree falls, it vibrates. When it vibrates, energy is transfered to the air. The air molecule will then vibrate and the so-called sound wave will then be transmitted into your ears.
512493
You musto to produces vibrations in your objeect, but you need a method to:\n\n1. producing vibration of differents frequencies\n2. A equiptment to measured the amplitud of the vibrations\n\nWhen the amplitud is big, you found the resonance frequency.\n\nYou should to observe that this maximum not occur at all frequencies, only a certain frequencies.\n\nOr you use a "stroboscopy light " to do it... produces the vibartion in your object and measure the vibration changing the stroboscopy frequency light.
32377
I wish your question was stated more clearly. Do you have sound data? From a microphone? \n\nOne of the things most often done is to estimate the spectral content of the data. If ou are inexperienced, which you appear to be, I would start with a standard direct periodogram estimate. \n\nSubtract the mean from your data. Take a discrete Fourier transform of the remaining data (it is easily done with the FFT algorithm, if the data are sampled at uniform intervals). Plot the modulus squared of the result. \n\nThere is a tremendous amount of sophisticated analyses done on sound samples. It is way beyond the scope of yahoo answers to be able to give much detail. You should do some research and reading in the library and on the web if you want more.
590541
Frequency 2Hz\n\nTime period = 1/2 s\n\nAmplitude = 10 cm.
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.
425520
I don't quite understand your notation, but here is how I would do this:\nThe general equation is y(x,t) = A sin 2 pi (t/T - x/L), where\nA = amplitude = 0.12 m\nL = wavelength = 0.3 m\nT = period\nThe latter is not given, but you can get it from the wavespeed v, via the formula v=L/T. Hence T = L/V = 0.3/6.4 = 0.0469 s.\nPlug A, L and T in the general equation and compare the result to the expression given in the problem.\nLooking at your expression, I would guess \na = A = 0.12 m \nb = 1/T = 21.3 s^-1\nc = 1/L = 3.33  m^-1