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Question
An electronically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of air column in the tube is 80 cm. The frequency of the loudspeaker can be varied between 20 Hz and 2 kHz. Find the frequencies at which the column will resonate. Speed of sound in air = 320 m s−1.
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Solution
Given:
Length of air column in the tube l = 80 cm = 80 × 10−2 m
Speed of sound in air v = 320 ms−1
The frequency of the loudspeaker can be varied between 20 Hz to 2 KHz.
The resonance column apparatus is equivalent to a closed organ pipe.
Fundamental note of a closed organ pipe is given by :
\[f = \frac{v}{4l}\]
\[\Rightarrow f = \frac{320}{4 \times 50 \times {10}^{- 2}} = 100 \text { Hz }\]
So, the frequency of the other harmonics will be odd multiples of f = (2n + 1)100 Hz.
According to the question, the harmonic should be between 20 Hz and 2 kHz.
∴ n = (0, 1, 2, 3, 4, 5, ..... 9)
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