First slide

Standing waves

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 $s^{- 1}$ )

Easy

A

4
B

5
C

7
D

6

Solution

Fundamental frequency of the closed organ pipe is
$v = \frac{v}{4 L}$
$Here, v = 340 m s^{- 1}, L = 85 cm = 0.85 m$
$∴ v = \frac{340 m s^{- 1}}{4 \times 0.85 m} = 100 Hz$
The natural frequencies of the closed organ pipe will be
$v_{n} = (2 n - 1) v = v, 3 v, 5 v, 7 v, 9 v, 11 v, 13 v, \dots$
$= 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz$
$1100 Hz, 1300 Hz, \dots and so on.$
Thus, the natural frequencies lies below the 1250 Hz is 6.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App