Two radio antennas are 10 km apart on a north-south axis on high mountain tops at the seacoast.The antennas broadcast identical AM radio signals,in phase,at a frequency of 4.70 MHz.A steamship,200 km offshore,travels toward the north at 15 km/h and passes east of the antennas.A radio on board the ship is tuned to the broadcast frequency.The reception of the radio signal on the ship is a maximum at a given instant.How much later until the next occurrence of maximum reception? (c = 3.00 × 10⁸ m/s) A)5.1 min B)3.8 min C)6.4 min D)7.7 min E)8.9 min

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Quizplus · Accepted Answer

To find the time until the next occurrence of maximum reception, we can use the concept of constructive interference from the two antennas. The path difference for constructive interference is given by $n\lambda$, where $n$ is an integer and $\lambda$ is the wavelength of the radio waves. The wavelength $\lambda$ can be calculated using the speed of light $c$ and the frequency $f$ of the radio waves: $\lambda = \frac{c}{f}$.Given:- $c = 3.00 \times 10^8 \, \text{m/s}$- $f = 4.70 \times 10^6 \, \text{Hz}$First, calculate the wavelength $\lambda$:$\lambda = \frac{3.00 \times 10^8 \, \text{m/s}}{4.70 \times 10^6 \, \text{Hz}} \approx 63.83 \, \text{m}$For maximum reception (constructive interference), the path difference should be an integer multiple of the wavelength. As the ship moves, the path difference changes due to the change in position relative to the antennas.The ship is moving parallel to the line connecting the two antennas, so the change in path difference as the ship moves northward is due to the change in the angle of reception. However, given the large distance (200 km offshore) compared to the separation of the antennas (10 km), this change in angle will be very small, and the primary factor for the change in path difference will be the direct northward movement of the ship.The next maximum occurs when the path difference increases by one wavelength, $\lambda$. Since the ship is moving at 15 km/h, we can calculate the time it takes for the ship to move a distance equal to one wavelength.Convert $\lambda$ to kilometers: $\lambda \approx 63.83 \, \text{m} = 0.06383 \, \text{km}$Time to move one wavelength: $t = \frac{\lambda}{\text{speed}} = \frac{0.06383 \, \text{km}}{15 \, \text{km/h}} \approx 0.004255 \, \text{h}$Convert time to minutes: $t \approx 0.004255 \, \text{h} \times 60 \, \text{min/h} \approx 0.2553 \, \text{min}$However, this calculation does not directly lead to the correct answer, indicating a mistake in the approach. The correct approach involves considering the phase difference caused by the additional path length traveled by the wave from one antenna to the ship compared to the other. The phase difference $\Delta \phi$ is related to the distance difference $\Delta d$ and the wavelength $\lambda$ by $\Delta \phi = \frac{2\pi \Delta d}{\lambda}$. For maximum reception, $\Delta \phi$ should change by $2\pi$, which corresponds to a change in path length of one wavelength, $\lambda$.Given the geometry and the specifics of the problem, the correct calculation involves understanding that the time until the next occurrence of maximum reception is determined by the difference in path lengths being an integer multiple of the wavelength, and as the ship moves, this condition is periodically met. The detailed calculation should account for the geometry of the situation and the speed of the ship, but the initial approach taken here missed the correct application of these principles to find the time until the next maximum reception. The correct answer involves a more detailed analysis of the interference pattern and the ship's motion relative to it.

Two Radio Antennas Are 10 Km Apart on a North-South