Two tuning forks are known to have frequency 258Hz and 264 hz and a maximum is observed just now, after how much time will the minimum be observed at the same plaece?(answer in second)

Explanation

To solve the problem, we need to find the time interval after which a minimum will be observed after a maximum when two tuning forks of frequencies 258 Hz and 264 Hz are played together.

### Step-by-Step Solution:

1. **Identify the Frequencies**:
- Let \( f_1 = 258 \, \text{Hz} \)
- Let \( f_2 = 264 \, \text{Hz} \)

2. **Calculate the Beat Frequency**:
- The beat frequency \( f_b \) is given by the absolute difference of the two frequencies:
\[ f_b = |f_2 - f_1| = |264 \, \text{Hz} - 258 \, \text{Hz}| = 6 \, \text{Hz} \]

3. **Calculate the Time Period of the Beat Frequency**:
- The time period \( T \) of the beat frequency is the reciprocal of the beat frequency:
\[ T = \frac{1}{f_b} = \frac{1}{6 \, \text{Hz}} = \frac{1}{6} \, \text{s} \approx 0.1667 \, \text{s} \]

4. **Determine the Time for Minimum**:
- When a maximum is observed, the next minimum will occur halfway through the time period of the beat frequency. Therefore, the time for the next minimum \( t_{\text{min}} \) is:
\[ t_{\text{min}} = \frac{T}{2} = \frac{1/6}{2} = \frac{1}{12} \, \text{s} \approx 0.0833 \, \text{s} \]

5. **Final Answer**:
- The time after which the minimum will be observed after the maximum is approximately:
\[ t_{\text{min}} \approx 0.083 \, \text{s} \]