can someone help me with sabines formula for reverberation time?
971302
here are some sites on the subject:
1133458
dB diff = 20 * log10(d1/d2) \n\n10 meters
1160585
4 Hertz = 4 times per second
510692
Try: allaboutphysics.com I think that might help.
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.
698460
Assume microwaves travel at the speed of light...\n\nv = 300,000,000 m/s\nwavelength(lambda) = 16 cm = .16 m\n\nFrequency = v/lambda\n\nFreq. = 300,000,000/.16\n\nFreq. = 1.875E9 Hz... or 1.875 GHz
304646
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
49259
c = yf\n\nc = speed of light (3 x 10^8 m/s)\ny = wavelength\nf = frequency\n\n==> f = c/y\n\n==> f = 3 x 10^8 m/s / 0.032 m \n\nf = 9.38 x 10^9 Hz\n\n-or-\n\nf = 9.38 GHz\n\n:)
716327
Goel has given you the correct answer.\n\nIf there is no damping of sound waves, sound once produced will last for ever. \n\nThe intensity of a sound wave having a frequency of 100 cycles per second, decreases by a factor of 1.015.\n\nThe intensity of a sound wave having frequency of 20, 000 cycles per second, decreases by a factor of 10^274.\n\nUltra sonic vibrations are damped so rapidly that their transmission over a distance of more than several hundred meters is completely impractical.\n\nWhen a source of sound in an auditorium is cut off the sound energy in the room does not vanish immediately. It takes certain time. This time depends upon the absorption of sound energy by the materials present in the room.\n\nThe persistence of sound in a room even after the source has been cut off is called reverberation.\n\nThe reverberation time is the time for the original intensity to fall to the audible limit.\n\nIf the reverberation time is below 0.5 second the speech is not heard in a room.\nIf the reverberation time is between 0.5 second and 3 second the speech is heard clearly.  \n \nIf the reverberation time is more than 3 second and if the time interval between consecutive reflections is less than 1/15 of a second, the sound is unclear.\n\nIf the time interval between consecutive reflections is more than 1/15 of a second, clear sound (echo) is heard.
752071
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.
355213
In y = A sin Bt, B is the frequency in radians/second, or B = 2pi*f, where f is the frequency in cycles/second, or Hertz.\n\nI assume by "c= 3 x 10^10", you're referring to the speed of light, which is 3*10^10 cm/sec.\n\nf = c / wavelength, so f = 3*10^10 cm/sec / (3*10^-10cm) = 3*10^20 cycles/sec or 3*10^20Hz.\n\nSo B = 2*pi*f = 2*pi*3*10^20 = 6*pi*10^20 radians/sec.  That's about 1.885*10^21 radians/sec.\n\nNote that these equations hold for all electromagnetic radiation, including radio waves, microwaves, IR, light, UV, X rays, and gamma rays.
887961
Use the general equation for Doppler effect from here( http://en.wikipedia.org/wiki/Doppler_effect ). f'/f = .86 and v0 =0 for you since the observer isn't moving. Note that since the frequency is lower that it must be moving away from you.
587882
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.
1093017
Your question is incomplete, part of it looks to be cut off.\n\nThis is just a guess, but the question almost seems like it is asking how long after the student claps his hands does he hear the eco bounce back off the wall.\nIn this case, one would need to know the speed of sound (or one could just estimate it to be about 340.29 m/s).\nSince speed is given as distance traveled per unit time, one can calculate the amount of time it would take something traveling at a given speed to be the distance traveled divided by the speed.  In this case with the sound, the sound traveled to the wall (165 meters) and back again (another 165 meters), for a total distance of 330 meters.
544479
The speed of sound: 340.29 m / s
44625
The speed of sound (at sea level) approximately is 340.29 m/s and can be used as a global estimate for the speed of sound for the purposes of this calculation.  That would mean that students measuring such a sound would need to be seperated by 340 meters to measure even a 1 second delay in sound propagation.  Additionally they would need to be able to perfectly synchronize the start of the sound event and the start of stopwatch event.  Making it even worse is the need to react with enough speed to stop the stopwatch upon reciept of the signal without significant reaction delay.\n\nThe experiment would be simply to accomplish with resources to control synchronization and detection equipment, but it would not be a low-budget, high-school type of experiment.
737785
I'll let you do the calculation, but the equation you need is:\n\nE = hc/lambda \n\nwhere h is Planck's constant, c is the speed of light in m/s and lambda is the wavelength of the light (in meters). You can find the rest of the numbers in your text book, I'm sure...