Problem 23 - University Physics with Modern Physics

Question

Points \(A\) and \(B\) are 56.0 \(\mathrm{m}\) apart along an east-west line. At each of these points, a radio transmitter is emitting a 12.5 -MHz signal horizontally. These transmitters are in phase with each other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 \(\mathrm{km}\) north of the \(A B\) line and initially placed at point \(C,\) directly opposite the midpoint of \(A B .\) The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain a range so that the intensity of the signal it receives from the transmitter is no less than \(\frac{1}{4}\) of its maximum value. How far from point \(C\) (along an east-west line) can the receiver be moved and always be able to pick up the signal?

Step-by-Step Solution

Answer

The maximum distance the receiver can be moved is found using constructive interference conditions.

Step 1: Find Wavelength of Signal

The frequency of the signal is given as 12.5 MHz. First, convert this frequency into Hertz, which is 12.5 x 10^6 Hz. The speed of light \(c\) is approximately \(3 \times 10^8 \mathrm{m/s}\). Use the formula \(\lambda = \frac{c}{f}\) to find the wavelength. Therefore, \(\lambda = \frac{3 \times 10^8}{12.5 \times 10^6} = 24 \mathrm{m}\).

Step 2: Establish Initial Conditions

Given the problem, the two transmitters at points \(A\) and \(B\) are separated by 56.0 m, and the receiver is initially positioned at 0.500 km (or 500 m) north of the midpoint of \(AB\). This means the path difference initially is zero when the receiver is at point \(C\).

Step 3: Determine Intensity Condition

The receiver can move such that the intensity is no less than 1/4 of the maximum intensity. The intensity of a wave is proportional to the square of the amplitude, so \(I = (\text{Amplitude})^2\). Therefore, if \(I = \frac{1}{4}I_0\), then the amplitude \(= \frac{1}{2}(\text{Amplitude}_0)\). This represents a condition of constructive or semi-constructive interference.

Step 4: Calculate Path Differences for Interference

For amplitudes to be \(\frac{1}{2}\) of the maximum, the path difference needs to be an odd multiple of \(\frac{\lambda}{2}\) (semi-constructive interference). Considering the geometry, the path difference from \(A\) and \(B\) to \(D\) where the condition satisfies is \(\frac{n\lambda}{2}\), where \(n = 1, 3, 5, \ldots\).

Step 5: Convert into Position Limits

Since the problem involves distance along an east-west line, use geometry to calculate the farthest distance the receiver can be moved. Assume a right-triangle scenario initially, where the hypotenuse changes as the receiver moves. The setup gives limitations on how far the distance can stretch away from \(C\) that still maintains the \(\frac{1}{4}\) intensity condition.

Step 6: Solve for Possible Maximum Distance

Using the setup, calculate the maximum distance the receiver can move. If necessary, employ trigonometric methods to resolve maximum path lengths into radial or east-west directions. The distance from the center line, \(x\), satisfies the equation for path, i.e., \(AB = \sqrt{(x^2 + 500^2)} - \sqrt{((56-x)^2 + 500^2)\) meeting optimal conditions.

Key Concepts

Radio Wave Propagation

Radio wave propagation refers to the movement or travel of radio waves through the atmosphere or space. This is an essential concept when discussing signals transmitted over distances, like between radio transmitters and receivers. In the exercise, two transmitters at points A and B emit signals in phase, meaning they start their wave cycles simultaneously. These waves travel horizontally, broadcasting in the same plane, creating interference patterns. Understanding how radio waves interact is about knowing these patterns and behaviors.

Consider that radio waves are electromagnetic waves, which means they travel at the speed of light, approximately 300,000,000 meters per second. The frequency of these waves, given in the problem as 12.5 MHz, helps determine how they propagate over various distances. As radio waves move, they might encounter obstacles or terrain changes, but, in this scenario, they uniformly spread in a horizontal plane, leading to interactions like interference.

Constructive Interference

Constructive interference occurs when two or more waves combine to form a wave of greater amplitude. This happens when waves align in phase, meaning their peaks (high points) and troughs (low points) match. In the exercise, the two in-phase transmitters create interference patterns that, depending on the position of the receiver, can amplify the signal.

For instance, at certain points along the receiver's path, the waves from transmitters A and B coincide constructively, resulting in a stronger received signal. This signal strength is tied to the amplitude of the waves. When the problem states that the receiver can receive an intensity of no less than one-quarter of the maximum intensity, it suggests points of constructive interference where the intensity remains strong. Such conditions allow the receiver to move within certain bounds along an east-west line while maintaining a clear signal reception.

Path Difference

Path difference refers to the difference in the distance traveled by two waves from their point of origin to a common point, like where a receiver detects them. In this problem, since the transmitters are 56.0 m apart, the path difference becomes crucial in understanding how their waves interact when reaching the receiver.The path difference determines whether the two radio waves interfere constructively or destructively at the receiver's location. For instance, when the path difference is a multiple of the wave's wavelength, constructive interference occurs, enhancing the signal's intensity. Conversely, if the path difference is a half-multiple of the wavelength, destructive interference may weaken or cancel the signal.

In the exercise, to maintain constructive or semi-constructive interference while moving away from point C, the path difference must only vary to satisfy conditions of \(\frac{n\lambda}{2}\), where \(n\) is an odd integer. This calculated path difference helps in determining the maximum distance the receiver can move along the east-west line while still receiving an intense enough signal.