For a closed organ pipe, three successive resonance frequencies are observed at 425 Hz, 595 Hz, and 765 Hz respectively. If the speed of sound in air is 340 m/s, then the length of the pipe is:
Correct answer: C. 1.0 m
- A. 2.0 m
- B. 0.4 m
- C. 1.0 m
- D. 0.2 m
Explanation
Given: v = 340 m/s, Ratio of 3 frequencies = 425 : 595 : 765Solution:Since frequencies are in odd number ratio, the pipe has to be a closed pipe.Ratio of 3 frequencies = 425 : 595 : 765 = 5 : 7 : 9So fundamental frequency= f = 425 / 5 = 85 HzFor fundamental frequencyl = v / 4f = 340 / (4 × 85) = 1 m.Options 2.0 m, 0.4 m, and 0.2 m are incorrect as they do not match the conditions and calculations for the given frequencies.
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Progressive waves transfer energy through oscillations, with wavelength, frequency, amplitude and wave speed related by v = fλ; sound speed depends on the medium. Superposition produces interference and stationary waves in strings and organ pipes, while simple harmonic motion describes the oscillation itself and must be distinguished from wave propagation.
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