A detector at `x=0` receives waves from three sources each of amplitude `A` and frequencies `f + 2,f` and `f-2`.
The equation of waves are : `y_(1)=A sin[2 pi(f+2),y_(2)=A sin 2 pift` and `y_(3)=A sin[2pi (f-2)t]`. The time at which intensity is minimum is

Options

A. `t=0,1//4,1//2,3//4,…sec`
B. `t=1//6,1//3,2//3,5//6,…sec`
C. `t=0,1//2,3//2,5//2,..sec`
D. `t=1//2,1//4,1//6,1//8,…sec`

Explanation

To find the time at which the intensity is minimum when three waves are received at a detector, we can follow these steps:

### Step 1: Write the equations of the waves
The equations of the waves from the three sources are given as:
- \( y_1 = A \sin(2 \pi (f + 2)t) \)
- \( y_2 = A \sin(2 \pi ft) \)
- \( y_3 = A \sin(2 \pi (f - 2)t) \)

### Step 2: Combine the wave equations
The total displacement \( y \) at the detector is the sum of the individual wave displacements:
\[ y = y_1 + y_2 + y_3 = A \sin(2 \pi (f + 2)t) + A \sin(2 \pi ft) + A \sin(2 \pi (f - 2)t) \]

### Step 3: Use the sine addition formula
We can use the sine addition formula to combine the waves. The formula states:
\[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
We can group the terms as follows:
\[ y = A \left( \sin(2 \pi (f + 2)t) + \sin(2 \pi (f - 2)t) \right) + A \sin(2 \pi ft) \]
Using the sine addition formula on the first two terms:
\[ y = A \left( 2 \sin\left(2 \pi ft\right) \cos(2 \pi t) \right) + A \sin(2 \pi ft) \]
This simplifies to:
\[ y = 2A \sin(2 \pi ft) \cos(2 \pi t) + A \sin(2 \pi ft) \]

### Step 4: Factor out common terms
Factoring out \( \sin(2 \pi ft) \):
\[ y = A \sin(2 \pi ft) (2 \cos(2 \pi t) + 1) \]

### Step 5: Find the intensity
The intensity \( I \) is proportional to the square of the amplitude:
\[ I \propto y^2 = A^2 \sin^2(2 \pi ft) (2 \cos(2 \pi t) + 1)^2 \]
To find the minimum intensity, we need to find when \( (2 \cos(2 \pi t) + 1)^2 \) is minimum.

### Step 6: Set the expression to zero
The expression \( (2 \cos(2 \pi t) + 1) = 0 \) gives:
\[ 2 \cos(2 \pi t) + 1 = 0 \implies \cos(2 \pi t) = -\frac{1}{2} \]

### Step 7: Solve for \( t \)
The cosine function is equal to \(-\frac{1}{2}\) at:
\[ 2 \pi t = \frac{2\pi}{3} + 2\pi n \quad \text{and} \quad 2 \pi t = \frac{4\pi}{3} + 2\pi n \]
Solving for \( t \):
\[ t = \frac{1}{3} + n \quad \text{and} \quad t = \frac{2}{3} + n \]
Where \( n \) is any integer.

### Step 8: Conclusion
The times at which the intensity is minimum are:
\[ t = \frac{1}{3}, \frac{2}{3}, \frac{5}{6}, \ldots \]