Two tunig forks of frequency `250Hz` and `256Hz` produce beats. If a maximum is observed just now, after how much time the next maximum is observed at the same place?

Options

A. `(1)/(18) s`
B. `(1)/(6) s`
C. `(1)/(12) s`
D. `(1)/(24) s`

Explanation

To solve the problem of determining the time until the next maximum is observed when two tuning forks of frequencies 250 Hz and 256 Hz produce beats, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Frequencies**:
- Let \( f_1 = 250 \, \text{Hz} \) (frequency of the first tuning fork)
- Let \( f_2 = 256 \, \text{Hz} \) (frequency of the second tuning fork)

2. **Calculate the Beat Frequency**:
- The beat frequency \( f_b \) is given by the absolute difference between the two frequencies:
\[ f_b = |f_2 - f_1| = |256 \, \text{Hz} - 250 \, \text{Hz}| = 6 \, \text{Hz} \]

3. **Determine the Time Period of the Beats**:
- The time period \( T_b \) of the beats is the reciprocal of the beat frequency:
\[ T_b = \frac{1}{f_b} = \frac{1}{6} \, \text{seconds} \]

4. **Calculate the Time Interval for Maximums**:
- The maximums in the beat pattern occur at intervals of half the beat period. Therefore, the time interval \( \Delta t \) between two consecutive maximums is:
\[ \Delta t = \frac{T_b}{2} = \frac{1/6}{2} = \frac{1}{12} \, \text{seconds} \]

5. **Conclusion**:
- Thus, if a maximum is observed just now, the next maximum will be observed after \( \frac{1}{12} \) seconds.

### Final Answer:
The next maximum will be observed at \( \frac{1}{12} \) seconds.

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