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Waves question

2014 · Q159
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Waves question

2014 · Q159

NEETPhysicsWavesMCQ+4 / −1
The number of possible natural oscillations of air column in a pipe closed at one end length 85 cm whose frequencies lie below 1250 Hz are (Velocity of sound = 340 m s−-−1)
  1. A
    4
  2. B
    5
  3. C
    7
  4. D
    6
View written solutionFree

Correct answer: D

Fundamental frequency of the closed organ pipe is

υ=v4L\upsilon = {v \over {4L}}υ=4Lv​

Here, v = 340 m s–1, L = 85 cm = 0.85 m

∴\therefore∴ υ=340 ms−14×0.85 m=100 Hz\upsilon = {{340\,m{s^{ - 1}}} \over {4 \times 0.85\,m}} = 100\,Hzυ=4×0.85m340ms−1​=100Hz

The natural frequencies of the closed organ pipe will be

υn=(2n−1)υ =υ,3υ,5υ,7υ,9υ,11υ....{\upsilon _n} = \left( {2n - 1} \right)\upsilon \, = \upsilon ,3\upsilon ,5\upsilon ,7\upsilon ,9\upsilon ,11\upsilon ....υn​=(2n−1)υ=υ,3υ,5υ,7υ,9υ,11υ....

= 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz, 1300 Hz,... and so on

Thus, the natural frequencies lies below the 1250 Hz is 6.

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