Given are two tuning forks near one another. One of them is of unknown frequency and the other is of frequency 591 Hz.We can hear beat of maximum intensity `I_(0)`with frequency 5 Hz. At t = 0 we hear a maxima. Then

Options

A. Unknown frequency can be 596 Hz
B. Unknown frequency can be 586 Hz
C. Intersity at time t = 2.7 s is /0
D. Intensity at time t = `(27)/(20)`s is` (0/2)`

Answer

To solve the problem step by step, we need to analyze the information given about the two tuning forks and the beat frequency.

### Step 1: Understand the concept of beat frequency
The beat frequency (\(f_B\)) is defined as the absolute difference between the frequencies of the two tuning forks. If one tuning fork has a known frequency (\(f_1 = 591 \, \text{Hz}\)) and the other has an unknown frequency (\(f_2\)), then:
\[ f_B = |f_2 - f_1| \]
Given that the beat frequency is \(5 \, \text{Hz}\), we can write:
\[ |f_2 - 591| = 5 \]

### Step 2: Set up equations for \(f_2\)
From the equation above, we can derive two possible equations for \(f_2\):
1. \(f_2 - 591 = 5\) (which gives us the maximum frequency)
2. \(591 - f_2 = 5\) (which gives us the minimum frequency)

### Step 3: Solve for \(f_2\)
1. For the first equation:
\[ f_2 = 591 + 5 = 596 \, \text{Hz} \]
2. For the second equation:
\[ f_2 = 591 - 5 = 586 \, \text{Hz} \]

Thus, the unknown frequency \(f_2\) can be either \(596 \, \text{Hz}\) or \(586 \, \text{Hz}\).

### Step 4: Analyze the intensity at a given time
At \(t = 0\), we hear a maxima, which indicates that the two frequencies are in phase. The intensity of the beats varies with time and is given by:
\[ I(t) \propto \cos^2(2\pi f_B t) \]
The maximum intensity occurs when \(t = 0\), and the intensity will drop to half its maximum value at specific intervals.

### Step 5: Determine the time for half intensity
The intensity is maximum at \(t = 0\) and will be half of the maximum intensity at certain times. The time at which this occurs can be calculated using the formula:
\[ \cos^2(2\pi f_B t) = \frac{1}{2} \]
This occurs at:
\[ 2\pi f_B t = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
For \(f_B = 5 \, \text{Hz}\):
\[ t = \frac{1}{2 \cdot 5} \left(\frac{1}{2} + n\right) = \frac{1}{20} + \frac{n}{10} \]

### Step 6: Evaluate the options
Now we can evaluate the options provided in the question based on the calculated frequencies and the behavior of intensity over time.

### Summary of Results
- The unknown frequency \(f_2\) can be either \(596 \, \text{Hz}\) or \(586 \, \text{Hz}\).
- The intensity at \(t = 0\) is maximum, and it will drop to half its maximum at specific intervals based on the beat frequency.