When two waves of nearly equal frequencies superimpose them. The frequency of combined wave is `'f_(1)'` that of the amplitude is `'f_(2)'` and the beat frequency is `'f_(3)'` Arrange them in increasing order of frequency

Options

A. `f_(1) lt f_(2) lt f_(3)`
B. `f_(3) lt f_(2) lt f_(1)`
C. `f_(2) lt f_(3) lt f_(1)`
D. `f_(1) lt f_(3) lt f_(2)`

Answer

To solve the problem of arranging the frequencies \( f_1 \), \( f_2 \), and \( f_3 \) in increasing order, we need to understand the definitions of each frequency in the context of wave superposition.

1. **Definitions**:
- \( f_1 \): Frequency of the combined wave (average frequency).
- \( f_2 \): Frequency of the amplitude (which is half the difference of the two frequencies).
- \( f_3 \): Beat frequency (the absolute difference between the two frequencies).

2. **Understanding the Frequencies**:
- Let's denote the two waves as having frequencies \( f_A \) and \( f_B \).
- The frequency of the combined wave, \( f_1 \), can be expressed as:
\[ f_1 = \frac{f_A + f_B}{2} \]
- The frequency of the amplitude, \( f_2 \), is given by:
\[ f_2 = \frac{|f_A - f_B|}{2} \]
- The beat frequency, \( f_3 \), is:
\[ f_3 = |f_A - f_B| \]

3. **Comparison of Frequencies**:
- Since \( f_3 \) is the absolute difference, it is always greater than or equal to \( f_2 \) (because \( f_2 \) is half of \( f_3 \)):
\[ f_2 < f_3 \]
- The average frequency \( f_1 \) is also less than or equal to the maximum of \( f_A \) and \( f_B \), thus:
\[ f_1 < f_3 \]
- Therefore, we can compare all three:
\[ f_2 < f_1 < f_3 \]

4. **Final Arrangement**:
- Arranging these in increasing order gives:
\[ f_2 < f_1 < f_3 \]

**Conclusion**: The increasing order of frequencies is \( f_2, f_1, f_3 \).

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