Sum

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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#### Solution

Given:

Length of organ pipe* \[L\]= 5 cm = 5 × 10 ^{−2} mv = 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 =\[\frac{20, 000}{3400} = 5 . 8 = 5\]

Concept: Speed of Wave Motion

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