The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (Velocity of sound = 340 m ) [2014]

Question

The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (Velocity of sound = 340 m $s^{- 1}$ ) [2014]

Smart Achievers · Accepted Answer

(4)

Fundamental frequency of the closed organ pipe is $υ = \frac{v}{4 L}$

Here, $v = 340$ $m$ $s^{- 1}$ , $L = 85$ $cm = 0.85$ $m$

$∴$ $υ = \frac{340 {m s}^{- 1}}{4 \times 0.85} = 100$ Hz

The natural frequencies of the closed organ pipe will be

$υ_{n} = (2 n - 1) υ = υ, 3 υ, 5 υ, 7 υ, 9 υ, 11 υ, 13 υ, \dots$

$= 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz, 1300 Hz, \dots$

and so on

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

Solution

Q.
The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (Velocity of sound = 340 m $s^{- 1}$ ) [2014]

1 4

2 5

3 7

4 6

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Solution

Q. The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (Velocity of sound = 340 m s-1) [2014]

1 4

2 5

3 7

4 6

Want Us to Email you About Special Offers & Updates?

Q.
The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (Velocity of sound = 340 m $s^{- 1}$ ) [2014]