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प्रश्न
A cylindrical metal tube has a length of 50 cm and is open at both ends. Find the frequencies between 1000 Hz and 2000 Hz at which the air column in the tube can resonate. Speed of sound in air is 340 m s−1.
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उत्तर
Given:
Length of cylindrical metal tube L = 50 cm
Speed of sound in air v = 340 ms−1
Fundamental frequency \[\left( f_1 \right)\]of an open organ pipe :
\[f_1 = \left( \frac{v}{2L} \right)\]
\[ = \frac{340}{2 \times 50 \times {10}^{- 2}} = 340 \text { Hertz }\]
So, the required harmonics will be in the range of 1000 Hz to 2000 Hz.
\[f_2 = 2 \times 340 = 680 \text { Hz }\]
\[ f_3 = 3 \times 340 = 1020 \text{ Hz }\]
\[ f_4 = 4 \times 340 = 1360 \text{ Hz }\]
\[ f_5 = 5 \times 340 = 1700 \text { Hz }\]
\[ f_6 = 6 \times 340 = 2040 \text { Hz }\]
f2, f3, f4... are the second, third, fourth overtone, and so on.
The possible frequencies between 1000 Hz and 2000 Hz are 1020 Hz, 1360 Hz and 1700 Hz.
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