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IIT JEE Advanced 2008 Paper 2
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© examsnet.com
Question : 2
Total: 66
A vibrating string of certain length l under tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces to 2 per second. Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz is
344
336
117.3
109.3
Validate
Solution:
The third harmonic of the closed pipe is
f =
3
V
4
l
=
3
×
340
4
×
0.75
= 340 Hz
This is also the frequency of the string since γ of the string is proportional to
√
T
. On increasing the tension, γ increases, and the number of beats decreases, given that n – 340 = 4. Therefore,
n > 340 Hz ⇒ n = 344 Hz
© examsnet.com
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