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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Apr 2007 11:23:20 IST
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Q) Calculate the least mass of air that would resonate witha tuning fork of frequency 340 Hz in a closed tube of diameter 4cm at 30 degree centigrade. (velocity of sound in air =350m/s;density of air at 30 degree centigrade=1.3kg per cube meter.) Neglect end correction.
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2 Apr 2007 11:58:14 IST
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is it 3.96 kg?
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light travels the fastest ??? NO
wherever light goes it always finds that darkness has already got there |
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2 Apr 2007 12:19:51 IST
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the ans is 0.42 gm
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2 Apr 2007 12:35:55 IST
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n = 340Hz & v = 350m/s
so,  = 350/340 m
min. length of air column would be in first harmonic
so, l = / 4 35/136 m
therefore, min. mass of air = V
 r2 l
(1.3 * 22 * .0004 * 35) / (136 * 7)
.42gm
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2 Apr 2007 12:42:06 IST
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hey punit "for minimum l = lambda/4 " this condtion is for tube open at both at the ends but how do we know whther it is iopen at both ends or closed at one end tube ?
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light travels the fastest ??? NO
wherever light goes it always finds that darkness has already got there |
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2 Apr 2007 13:07:21 IST
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rajat, read the ques. again, its given that tube is closed
l = lambda / 4
this condition is for tube open at one end & closed at other
actually, in the question it is not mentioned that whether the tube is closed at both ends or at one end
But, i guessed that it is open at one end as the last line says ' neglect end correction'
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2 Apr 2007 13:30:03 IST
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punit , i think L = (n - 1/2)  /2 is the condition for tube open at one end and closed at another whereas L = n /4 is the condition for tube open at "both " ends . see H.C verma ,it is given clearly in it . and now,if u say it is closed at one end then why did u apply the second equation which is for open at both ends ???
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3 Apr 2007 12:54:11 IST
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rajat plz open pg. no. 338 of HCV part-i and then pg 337
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