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Natural Science Forum / Physics / Acoustics / March 2006Speed of sound in relation to temprature - NEWBIE()!!!!
Hi Guys, Please note the newbie above!!! I've start looking at physics and am a mature student and on the course I'm doing we were given the follownig questions and answers, but no explantation. Are the answers below correct? Question Given that sound speed will increase at a set rate for every 1 degree increase in temperature. Assuming we have the following temperatures: (is this even correct - Sarah) 0C = 332M/Sec -10C = 323 M/Sec +30 = 352 M/Sec 1. Based above, what is the rate of increase for 1 degree C? 3 2. Based on the previous, what is the speed of sound at -37.5C? Answers: 1: -10 => 323 M/Sec +30 => 352 M/Sec =>Difference in temperature is 40 degrees C Difference in speed is = 352 - 323 = 29 M/Sec => speed (29) / temperature (40) = 0.725 => the speed of sound increases by 0.725M/Sec for every 1 degree rise in temperature 2. Based on the previous example, calculate the speed of sound at -37.5 degrees. We know that -10 => 323 M/Sec +30 is 352 -37.5C is 27.5 lower => difference of 27.5 Multiply 27.5 * 0.725 => 19.9375 As the air temperature is negative we subtract => 323 - 19.9375 => 303.0265 M/Sec Is the above remotely right? For Q2, why is 19.9375 subtracted from the speed of sound at -10, is it because in the Question that was the lowest temprature quotred? If you know the answer, please explain in plain English. I've checked the net but the stuff is way over my head or is there another way the answer question 1 and 2? Any help would be greatly appreciated. Regards, Sarah Hamilton. > I've start looking at physics and am a mature student and on the > course [quoted text clipped - 34 lines] > answer question 1 and 2? > Any help would be greatly appreciated. That says it all: http://www.sengpielaudio.com/calculator-speedsound.htm http://www.sengpielaudio.com/SpeedOfSoundPressure.pdf http://www.sengpielaudio.com/calculator-airpressure.htm Wikipedia helps: http://en.wikipedia.org/wiki/Speed_of_sound An approximate speed (in metres per second) can be calculated from: Speed of sound c = 331.5 + 0.6 * theta where theta is the temperature in degrees Celsius. That says 1 degree changes 0.60 m in sound speed. -37.5°C = 307.86 m/s +30°C = 348.18 m/s difference is 67.5°C = 40.32 m/s that is 0.60 m per degree. Cheers Jens Thanks for the info., I'll check out the sites So the stuff I put in above is completely wrong as regards maths? Regards, Sarah Hamilton. > Thanks for the info., I'll check out the sites. > So the stuff I put in above is completely wrong as regards maths? Yes, when you get a different answer from speed of sound c = 331.5 + 0.6 * theta, where theta is the temperature in degrees Celsius. 1 degree changing of temperature is equal to 60 cm changing of speed of sound. Regards Eberhard Sengpiel German forum for microphone recordings and sound studio techniques http://www.sengpielaudio.com/Calculations03.htm > An approximate speed (in metres per second) can be calculated from: > Speed of sound c = 331.5 + 0.6 * theta snip. Another approximation (requires square root computation) is: US c= 49*SQRT(459+F) ft/sec. F=fahrenheit temp SI c=20*SQRT(273+C) m/s. C=Centrigrade temp. Angelo Campanella [snip] >Question >Given that sound speed will increase at a set rate for every 1 degree >increase in temperature. Assuming we have the following temperatures: [snip] You've been given a math problem on how to do linear interpolation/extrapolation. The speed of sound in air varies as the square root of absolute temperature. You approximate it as a linear relation over small temperature ranges. But once you do that you've lost touch with the physics - it's just math. On the math side, life would seemingly be easier if you take two of the points > 0C = 332M/Sec > -10C = 323 M/Sec > +30 = 352 M/Sec and fit an equation V = a + bT Notice that the three points you are given is not consistent with the assertion that there is a set rate per degree. If you take the first two points, the rate (b in my equation) is 0.9 M/sec/deg. If you take the first and third, the rate is 0.67 M/sec/deg. Are you sure you've re-stated your homework problem correctly? Ken Plotkin > Hi Guys, > [quoted text clipped - 49 lines] > > Any help would be greatly appreciated. Usually it is best to try to fit data with an equation that has some theoretical basis (and therefore some relationship between the data and the formula). It shows in http://en.wikipedia.org/wiki/Speed_of_sound and http://scienceworld.wolfram.com/physics/SpeedofSound.html that the speed of sound in air is proportional to the *square root* of absolute temperature. v = k * sqrt(T) If we analyze your data this way (plotting speed in meters per second against temperature in a linear regression - I used Excel) we get v = 23.99 * sqrt(T(K)) - 65.48 where the meaning of the coefficients is not so important as the fact that this gives a really good fit to the data, and the following temp data calc T(K) -10 323 323.7 263.16 0 332 331.0 273.16 +30 352 352.2 303.16 Applying the same line to your point of interest (-37.5 C = 235.66 K) gives a v = 302.8 m/sec. Please understand that 'fits' to data are generally OK for *interpolation* (i.e. calculating values intermediate to those measured) but extrapolations are chancy, and become less reliable the further one gets from the midpoint of the data range. This number suggests that your number is *real* close to right. Tom Davidson Richmond, VA |
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