Time interval for which the detector remains idle in each cycle of intensity

Options

A. `1/3`sec
B. `2/3`sec
C. `1/2`sec
D. `1/4`sec

Explanation

To solve the problem of determining the time interval for which the detector remains idle in each cycle of intensity, we can follow these steps:

### Step 1: Understand the Problem
We have two sound sources with frequencies \( f_1 = 400 \, \text{Hz} \) and \( f_2 = 401 \, \text{Hz} \). The detector can only detect sound when the amplitude of the resultant wave is greater than \( A \).

### Step 2: Write the Equations for the Waves
The equations for the waves emitted by the two sources can be written as:
- For source 1: \( y_1 = A \sin(2\pi f_1 t) \)
- For source 2: \( y_2 = A \sin(2\pi f_2 t) \)

### Step 3: Find the Resultant Amplitude
The resultant amplitude when both waves interfere can be expressed as:
\[ A_{\text{resultant}} = 2A \cos\left(\pi (f_1 - f_2) t\right) \]
Given \( f_1 - f_2 = -1 \, \text{Hz} \), we can simplify this to:
\[ A_{\text{resultant}} = 2A \cos(\pi t) \]

### Step 4: Set the Condition for Detection
The detector will only detect the sound when the amplitude is greater than \( A \). Thus, we set up the inequality:
\[ 2A \cos(\pi t) > A \]
Dividing both sides by \( A \) (assuming \( A > 0 \)):
\[ 2 \cos(\pi t) > 1 \]
This simplifies to:
\[ \cos(\pi t) > \frac{1}{2} \]

### Step 5: Solve for \( t \)
The condition \( \cos(\pi t) > \frac{1}{2} \) corresponds to the angles where the cosine function is greater than \( \frac{1}{2} \). The angles satisfying this condition are:
\[ -\frac{\pi}{3} < \pi t < \frac{\pi}{3} \quad \text{and} \quad \frac{5\pi}{3} < \pi t < \frac{7\pi}{3} \]
Dividing through by \( \pi \):
\[ -\frac{1}{3} < t < \frac{1}{3} \quad \text{and} \quad \frac{5}{3} < t < \frac{7}{3} \]

### Step 6: Determine the Idle Time
The detector is idle when \( \cos(\pi t) \leq \frac{1}{2} \):
1. From \( t = \frac{1}{3} \) to \( t = \frac{5}{3} \), the detector is idle.
2. The time interval for which the detector is idle is:
\[ \Delta t = \frac{5}{3} - \frac{1}{3} = \frac{4}{3} \, \text{seconds} \]

### Final Result
The time interval for which the detector remains idle in each cycle of intensity is \( \frac{4}{3} \, \text{seconds} \).

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