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प्रश्न
An open organ pipe has a length of 5 cm. (a) Find the fundamental frequency of vibration of this pipe. (b) What is the highest harmonic of such a tube that is in the audible range? Speed of sound in air is 340 m s−1 and the audible range is 20-20,000 Hz.
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उत्तर
Given:
Length of organ pipe \[L\]= 5 cm = 5 × 10−2 m
v = 340 m/s
The audible range is from 20 Hz to 20,000 Hz.
The fundamental frequency of an open organ pipe is : \[f = \frac{v}{2L}\]
On substituting the respective values ,we get : \[f = \frac{340}{2 \times 2 \times {10}^{- 2}} = 3 . 4 \text { kHz }\]
(b) If the fundamental frequency is 3.4 kHz, then the highest harmonic in the audible range (20 Hz - 20 kHz) is
Required highest harmonic =\[\frac{20, 000}{3400} = 5 . 8 = 5\]
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