If f1 and f2 are frequencies of two tuning forks, such that f1 < f2 , then number of beats produced in one second are:
Correct answer: A. f2-f1
- A. f2-f1
- B. f2 * fl
- C. f1 - f2
- D. f2 + f1
Explanation
The number of beats produced in one second when two tuning forks with frequencies f1 and f2 (where f1 < f2) are sounded together can be calculated using the formula:Number of beats = |f1 - f2|Here, |x| represents the absolute value of x. The absolute value ensures that the number of beats is always positive, regardless of which frequency is higher.So, the number of beats produced in one second would be |f1 - f2|.As this is a numerical so only one option is correct.
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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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