If two sound waves with frequencies of 128 Hz and 126 Hz are produced simultaneously by two different tunning forks, then the detected beat will have a frequency
Correct answer: A. 2 Hz.
- A. 2 Hz.
- B. 126 Hz.
- C. 128 Hz.
- D. 256 Hz.
Explanation
When two sound waves with close frequencies are played together, they produce beats. The beat frequency is calculated by taking the absolute difference between the two frequencies. In this case, the frequencies are 128 Hz and 126 Hz, so the beat frequency is |128 Hz - 126 Hz| = 2 Hz. Option A is correct. Option B and Option C, 126 Hz and 128 Hz, respectively, are the original frequencies and not the beat frequency. Option D, 256 Hz, is the sum of the two frequencies, which is not relevant in calculating the beat frequency.
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About Superposition and Interference
Superposition states that overlapping waves produce a resultant displacement equal to the algebraic sum of their individual displacements. Interference covers constructive and destructive interference, phase difference, path difference, coherent sources, and conditions for maxima and minima. It differs from diffraction, which involves spreading around openings or obstacles.
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