If `n_(1) and n_(2)` are the frequencies of sound waves producing beats then the time interval between one maxima and next minima is

Options

A. `(1)/(n_(1)+n_(2))`
B. `(1)/(n_(1)-n_(2))`
C. `(1)/(2(n_(1)+n_(2)))`
D. `(1)/(2(n_(1)-n_(2)))`

Answer

To solve the problem of finding the time interval between one maxima and the next minima when two sound waves with frequencies \( n_1 \) and \( n_2 \) produce beats, we can follow these steps:

### Step 1: Understand the concept of beats
When two sound waves of slightly different frequencies interfere, they produce a phenomenon known as beats. The beat frequency is given by the absolute difference of the two frequencies:
\[ f_{\text{beat}} = |n_1 - n_2| \]

### Step 2: Determine the time period of beats
The time period \( T \) of the beat is the reciprocal of the beat frequency:
\[ T = \frac{1}{f_{\text{beat}}} = \frac{1}{|n_1 - n_2|} \]

### Step 3: Relate maxima and minima to beats
In the context of beats, one complete cycle (or one beat) consists of one maximum and one minimum. Therefore, the time interval for one complete beat (from one maximum to the next maximum) is \( T \).

### Step 4: Calculate the time interval between one maximum and the next minimum
Since one beat consists of one maximum and one minimum, the time interval between one maximum and the next minimum is half of the time period of one beat:
\[ \text{Time interval between one maximum and next minimum} = \frac{T}{2} = \frac{1}{2|n_1 - n_2|} \]

### Step 5: Final expression
Thus, the time interval between one maxima and the next minima can be expressed as:
\[ \text{Time interval} = \frac{1}{2(n_1 - n_2)} \quad \text{(assuming } n_1 > n_2\text{)} \]

### Conclusion
The correct answer to the question is:
\[ \text{Time interval between one maxima and next minima} = \frac{1}{2(n_1 - n_2)} \]