A pipe closed at one end has length 83 cm. The number of possible natural oscillations of air column whose frequencies lie below 1000 Hz are (velocity of sound in air = 332 m/s)

Question

A pipe closed at one end has length 83 cm. The number of possible natural oscillations of air column whose frequencies lie below 1000 Hz are (velocity of sound in air = 332 m/s) (A) 3 (B) 4 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

Fundamental frequency of one end closed pipe,

v = \frac{n v}{4 L}

Here,

n = 1

or

v_{1} = \frac{332}{4 \times 83 \times 1 0 ^{- 2}}

or

v_{1} = 100

As odd harmonics alone are produced in closed organ pipe, therefore possible frequencies are

3 v_{1} = 300 Hz, 5 v_{2} = 500 Hz

7 v_{2} = 700 Hz, 9 v_{2} = 900 Hz

Hence, the number of possible natural oscillation of air column is

5.

Q. A pipe closed at one end has length 83cm. The number of possible natural oscillations of air column whose frequencies lie below 1000Hz are (velocity of sound in air =332m/s)

3

4

5

6

Q. A pipe closed at one end has length $83 c m$ . The number of possible natural oscillations of air column whose frequencies lie below $1000 Hz$ are (velocity of sound in air $= 332 m / s$ )