GR0177 #50


Problem


This problem is still being typed. 
Wave Phenomena}Sound
Since the wavelength of the wave does not change, as the pipe presumably stays approximately the same length, only the frequency varies. If the speed of sound changes, then the frequency changes. If the speed of sound is lower than usual, then the frequency is lower. Thus, choices (A), (B) and (C) remain. Calculate to get choice (B).


Alternate Solutions 
uhurulol 20141020 18:24:42  If you're looking for a more mathematical answer, use the simple relation
=
Note that is just , and as such
= .
cancels on both sides and , so plugging everything in and cancelling we have
Cross multiply to obtain , as in choice (B). Hope this helps.   dham 20101006 23:07:59  For a quick guess: Cold instruments go flat. (Frequency goes down), leaving choices AC.
Directly: c=f, so speed of sound is directly proportional to wavelength.
Find the change in wavelength by finding (440)=3*4.4.
That's little over 12.
44012=428, which is a little over 427 (B).
B is the right answer.  

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  yummyhat 20171026 12:40:36  use the fact that wavelength*frequency=velocity. notice that the wavelength is determined by the pipe and is constant. use a ratio because the ratio of velocities is already given, and solve for the final frequency.   kevintah 20151005 10:05:11  Hey guys, . . . . Kevin Tah Njokom here :)\r\n\r\n Currently studying for the Physics GRE. This is my first response here. I think it is how to attack the problem, although you have to be able to do it fast to account for time. So here goes.\r\n First a few observations. I noticed that the solution was given by @uhurulol using some very quick method, which is probably faster. My solution is an alternative way of arriving at the answer.\r\n\r\n Notice that the pipe is open at one end. This makes the following relation valid. . Here L is the length of the pipe . This is a good thing because we know that the length is not suddenly going to change. \r\n We also know the relation that . \r\n We solve for from this, and then substitute the value into the equation for . This gives us . Since is constant it mus apply for the new and old frequency and velocity.\r\n\r\n It is mentioned that on a cold day, the speed of sound is 3% lower than it would be at 20 degrees.\r\n\r\n This means \r\n \r\nThis makes \r\n\r\nThus:\r\n\r\n \r\n\r\nWhen you crunch things out, you get 426.8 close to 427. This makes B, the answer.\r\n\r\n Good Luck on the Physics GRE everyone. Study hard. Let me know what you think of my answer.
kevintah 20151005 23:39:49 
Some how the Latex got distorted, and I can\'t edit it . Oh well

  uhurulol 20141020 18:24:42  If you're looking for a more mathematical answer, use the simple relation
=
Note that is just , and as such
= .
cancels on both sides and , so plugging everything in and cancelling we have
Cross multiply to obtain , as in choice (B). Hope this helps.   jwbrooks0 20140922 21:21:30  I'm confused. The speed of sound should be proportional to the sqrt(T). http://en.wikipedia.org/wiki/Speed_of_sound#Speed_in_ideal_gases_and_in_air
Therefore, why isn't 440*sqrt(0.97) = 433 correct?
Stardust 20151022 15:37:18 
The question says the \'SPEED OF SOUND is 3% lower than...\' not the temperature.

  dham 20101006 23:07:59  For a quick guess: Cold instruments go flat. (Frequency goes down), leaving choices AC.
Directly: c=f, so speed of sound is directly proportional to wavelength.
Find the change in wavelength by finding (440)=3*4.4.
That's little over 12.
44012=428, which is a little over 427 (B).
B is the right answer.   f4hy 20091106 23:27:57  They give you the 20 degrees as a trick then?
dham 20101006 23:09:25 
Not really, it's just an initial condition.

  michealmas 20061230 11:48:37  use the equation relating freq., velocity, and wavelength.   nahmad 20060330 22:21:27  One quick way of doing this calculation in your head is to consider it as a 3% loss. Thus 3 for every 100, which gives a loss of 12. 44012 = 428.  

Post A Comment! 
You are replying to:
Hey guys, . . . . Kevin Tah Njokom here :)\r\n\r\n Currently studying for the Physics GRE. This is my first response here. I think it is how to attack the problem, although you have to be able to do it fast to account for time. So here goes.\r\n First a few observations. I noticed that the solution was given by @uhurulol using some very quick method, which is probably faster. My solution is an alternative way of arriving at the answer.\r\n\r\n Notice that the pipe is open at one end. This makes the following relation valid. . Here L is the length of the pipe . This is a good thing because we know that the length is not suddenly going to change. \r\n We also know the relation that . \r\n We solve for from this, and then substitute the value into the equation for . This gives us . Since is constant it mus apply for the new and old frequency and velocity.\r\n\r\n It is mentioned that on a cold day, the speed of sound is 3% lower than it would be at 20 degrees.\r\n\r\n This means \r\n \r\nThis makes \r\n\r\nThus:\r\n\r\n \r\n\r\nWhen you crunch things out, you get 426.8 close to 427. This makes B, the answer.\r\n\r\n Good Luck on the Physics GRE everyone. Study hard. Let me know what you think of my answer.

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