Two tuning fork having frequency 300 Hz & 305 Hz produce beat phenomena. Then (i) How many beats produce is 5 sec. (ii) Determine minimum time interval in which maximum intensity become min.

Explanation

To solve the problem step by step, we will break it down into two parts as per the questions asked.

### Part (i): Number of Beats Produced in 5 Seconds

1. **Identify the Frequencies**:
We have two tuning forks with frequencies:
- \( f_1 = 300 \, \text{Hz} \)
- \( f_2 = 305 \, \text{Hz} \)

2. **Calculate the Beat Frequency**:
The beat frequency \( f_b \) is given by the absolute difference between the two frequencies:
\[ f_b = |f_2 - f_1| = |305 \, \text{Hz} - 300 \, \text{Hz}| = 5 \, \text{Hz} \]

3. **Calculate the Number of Beats in 5 Seconds**:
The number of beats produced in a given time can be calculated using the formula:
\[ \text{Number of beats} = f_b \times t \]
where \( t \) is the time in seconds. For 5 seconds:
\[ \text{Number of beats} = 5 \, \text{Hz} \times 5 \, \text{s} = 25 \, \text{beats} \]

### Part (ii): Minimum Time Interval for Maximum Intensity to Minimum Intensity

1. **Understanding Maximum and Minimum Intensity**:
The maximum intensity occurs when the amplitudes of the waves are in phase (constructive interference), and minimum intensity occurs when they are out of phase (destructive interference).

2. **Determine the Beat Frequency**:
We already calculated the beat frequency \( f_b = 5 \, \text{Hz} \).

3. **Calculate the Time Period of the Beats**:
The time period \( T \) of the beat frequency is given by:
\[ T = \frac{1}{f_b} = \frac{1}{5 \, \text{Hz}} = 0.2 \, \text{s} \]

4. **Find the Minimum Time Interval**:
The minimum time interval from maximum intensity to minimum intensity is half of the time period:
\[ \text{Minimum time interval} = \frac{T}{2} = \frac{0.2 \, \text{s}}{2} = 0.1 \, \text{s} \]

### Final Answers:
(i) The number of beats produced in 5 seconds is **25 beats**.
(ii) The minimum time interval in which maximum intensity becomes minimum is **0.1 seconds**.