Two radio stations broadcast their programmes at the same amplitude A and at slightly different freq

Q: Why is the intensity proportional to $4A^2$ instead of just $2A^2$ at its maximum?

When two coherent waves of amplitude $A$ interfere constructively, their amplitudes add up to $2A$. Since intensity is proportional to the square of the amplitude, the maximum intensity is proportional to $(2A)^2 = 4A^2$.

Q: Why does the period of the intensity cycle correspond to $\theta$ changing by $\pi$ instead of $2\pi$?

The intensity depends on $\cos^2\theta$. While $\cos\theta$ has a period of $2\pi$, squaring it halves the period to $\pi$ because the negative half-cycles of the cosine wave become positive when squared.

Q: What does it mean for the detector to be 'idle'?

The detector has a minimum threshold of intensity ($2A^2$) below which its circuitry cannot register any signal. During these times, the detector behaves as if no signal is present, hence it is 'idle'.

Q: How would the idle time change if the threshold intensity was increased?

If the threshold intensity is increased, the detector would require a higher intensity to activate. This would expand the range of angles where the intensity is below the threshold, thereby increasing the idle time.

Q: Is the beat frequency equal to $\omega_1 - \omega_2$ or $(\omega_1 - \omega_2)/2$?

The beat frequency of the amplitude envelope is $(\omega_1 - \omega_2)/2$, but because intensity depends on the square of the amplitude, the intensity peaks twice as often. Thus, the frequency of the intensity beats is exactly $\omega_1 - \omega_2$.

Question

Two radio stations broadcast their programmes at the same amplitude $A$ and at slightly different frequencies $\omega_1$ and $\omega_2$ respectively, where $\omega_1 - \omega_2 = 10^3 	ext{ Hz}$. A detector receives the signals from the two stations simultaneously. It can only detect signals of intensity $\ge 2 A^2$.

(a) Find the time interval between successive maxima of the intensity of the signal received by the detector.
(b) Find the time for which the detector remains idle in each cycle of the intensity of the signal.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing between two powerful radio transmitters broadcasting at slightly different frequencies.

As the waves travel through space and reach your detector, they don't just pass by independently; they **interfere** with each other.

This interference is the physical origin of the phenomenon known as **beats**.

Let's write down the mathematical expressions for these two individual waves:

$$y_1 = A \sin(\omega_1 t)$$

$$y_2 = A \sin(\omega_2 t)$$

Here, both waves have the same amplitude $A$, but their angular frequencies $\omega_1$ and $\omega_2$ differ slightly.

### The Principle of Superposition

When these two waves arrive at the detector simultaneously, the net displacement $y$ is the algebraic sum of the individual displacements:

$$y = y_1 + y_2 = A \sin(\omega_1 t) + A \sin(\omega_2 t)$$

To simplify this sum, we can use the standard trigonometric identity:

$$\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}ight) \cos\left(\frac{C-D}{2}ight)$$

Applying this identity to our superposition equation yields:

$$y = 2A \cos\left(\frac{\omega_1 - \omega_2}{2} tight) \sin\left(\frac{\omega_1 + \omega_2}{2} tight)$$

This is a beautiful result! It represents a high-frequency wave (with frequency $\frac{\omega_1 + \omega_2}{2}$) whose amplitude is slowly modulated by a envelope function:

$$A_R(t) = 2A \cos\left(\frac{\omega_1 - \omega_2}{2} tight)$$

### Finding the Intensity Profile

The intensity $I$ of a wave is directly proportional to the square of its amplitude. Therefore, the time-dependent intensity of the combined signal is:

$$I(t) \propto A_R^2(t) \implies I(t) = 4A^2 \cos^2\left(\frac{\omega_1 - \omega_2}{2} tight)$$

We are given that the frequency difference is:

$$\omega_1 - \omega_2 = 10^3 	ext{ s}^{-1}$$

Substituting this value into our intensity equation gives:

$$I(t) = 4A^2 \cos^2(500 t)$$

### Part (a): Time Interval Between Successive Maxima

The intensity reaches its maximum value of $4A^2$ when the cosine squared term is equal to $1$:

$$\cos^2(500 t) = 1 \implies 500 t = n\pi$$

Solving for $t$, we get the times at which maxima occur:

$$t_n = \frac{n\pi}{500} \quad 	ext{for } n = 0, 1, 2, \dots$$

The time interval $\Delta t$ between any two successive maxima is simply:

$$\Delta t = t_{n+1} - t_n = \frac{\pi}{500} 	ext{ s}$$

Using the approximation $\pi \approx 3.1416$:

$$\Delta t \approx 6.28 	imes 10^{-3} 	ext{ s} = 6.28 	ext{ ms}$$

### Part (b): Finding the Idle Time of the Detector

The detector has a physical limitation: it can only register a signal if the intensity is greater than or equal to $2A^2$.

Therefore, the detector remains **idle** whenever the intensity drops below this threshold:

$$I(t) < 2A^2 \implies 4A^2 \cos^2(500 t) < 2A^2$$

Dividing both sides by $4A^2$ simplifies this to:

$$\cos^2(500 t) < \frac{1}{2}$$

Let's define the phase angle $	heta = 500 t$. The idle condition becomes:

$$|\cos	heta| < \frac{1}{\sqrt{2}}$$

Let's analyze one full cycle of the intensity variation, which corresponds to $	heta$ spanning from $0$ to $\pi$ (since $\cos^2	heta$ has a period of $\pi$):

* The detector is **active** when $|\cos	heta| \ge \frac{1}{\sqrt{2}}$, which corresponds to:
  $$	heta \in \left[0, \frac{\pi}{4}ight] \cup \left[\frac{3\pi}{4}, \piight]$$

* The detector is **idle** when $|\cos	heta| < \frac{1}{\sqrt{2}}$, which corresponds to:
  $$	heta \in \left(\frac{\pi}{4}, \frac{3\pi}{4}ight)$$

The angular width of this idle phase in each cycle is:

$$\Delta	heta_{	ext{idle}} = \frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2}$$

Since $	heta = 500 t$, the actual time duration $t_{	ext{idle}}$ for which the detector remains idle in each cycle is:

$$t_{	ext{idle}} = \frac{\Delta	heta_{	ext{idle}}}{500} = \frac{\pi/2}{500} = \frac{\pi}{1000} 	ext{ s}$$

Using the approximation $\pi \approx 3.1416$:

$$t_{	ext{idle}} \approx 3.14 	imes 10^{-3} 	ext{ s} = 3.14 	ext{ ms}$$

This means that in every cycle of $6.28 	ext{ ms}$, the detector is active for $3.14 	ext{ ms}$ and idle for exactly $3.14 	ext{ ms}$!

Visualized Solution

Visualizing the Superposition Setup

Applying the Principle of Superposition

Extracting the Amplitude and Intensity

Substituting the Frequency Difference

Part (a): Finding the Time Interval Between Maxima

Part (b): Setting Up the Idle Condition

Solving the Trigonometric Inequality

Calculating the Idle Time Interval

Final Summary of Results

The Sigma Insight: Interference of Waves

Analyzing the Setup

The Principle of Superposition

Finding the Intensity Profile

Part (a)

Part (b)

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