Question 489: Waves

Question

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)

Accepted Answer

Fundamental frequency of the closed organ pipe is
$$\upsilon = {v \over {4L}}$$
Here, v = 340 m s
Ã¢ÂÂ1
, L = 85 cm = 0.85 m
$$ 	herefore $$ $$\upsilon  = {{340\,m{s^{ - 1}}} \over {4 	imes 0.85\,m}} = 100\,Hz$$
The natural frequencies of the closed organ pipe will be
$${\upsilon _n} = \left( {2n - 1} ight)\upsilon \, = \upsilon ,3\upsilon ,5\upsilon ,7\upsilon ,9\upsilon ,11\upsilon ....$$
= 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.

Answer

4

Answer

5

Answer

7

Answer

6

Waves

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)

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