Calculating radio station frequency given distances

In summary, the problem involves estimating the frequency of a radio station by using the interference pattern created by two transmission towers. Using the given values of the towers' distance and the distance from the present location, the wavelength and frequency can be calculated using the formula for two source interference.
  • #1
aeliand
1
0
Sorry if I posted this in the wrong location, it seemed like the most appropriate board.

Homework Statement


As you drive down the highway, you notice that the dial on your stereo is not functioning. You have the radio tuned to a station that uses two transmission towers that are 175.0 m apart. The towers are 25.00 km from your present location. You wish to estimate the frequency setting of your radio using the interference pattern set up by two towers. You notice that the signal reception fluctuated between maximums as you drive a distance of 0.4500 km parallel to the line joining the two towers. What is the frequency of the station to which you are listening?

Homework Equations


This is supposed to test at how well we can find out data about upcoming units, so I had to guess at which equations would be most relevant since we have yet to learn them. This formula is the only one I know of that I figured would help in any way.

velocity = frequency * lambda

The Attempt at a Solution


I assumed by maximums, it means 0.45/2 on each side of the "present location", so I calculated the distance to the points from each tower.

I drew out the diagram, which made an isoceles triangle with dimensions 0.1750 km, 25 km and 25 km, then split the triangle in half to figure out the distance between the line created by the two receivers and the highway, and using Pythagorean theorem I got a value around 24.99984687 km. I then used that value to find out the distance from the maximums to the towers. From either point to the tower farthest from it, it would be 25.00179994 km away, and to the tower closest it would be 25.000225 km.

I calculated the time to reach each tower, and the values were 0.073520961 s to the closest tower and 0.073534705 s to the farthest.

I assume from here it would do something with resonance but I can not think of the correct formula to use to calculate that. I don't necessarily need a full answer, but a step in the right direction would be very helpful. Thanks in advance!
 
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  • #2
aeliand said:
Sorry if I posted this in the wrong location, it seemed like the most appropriate board.

Homework Statement


As you drive down the highway, you notice that the dial on your stereo is not functioning. You have the radio tuned to a station that uses two transmission towers that are 175.0 m apart. The towers are 25.00 km from your present location. You wish to estimate the frequency setting of your radio using the interference pattern set up by two towers. You notice that the signal reception fluctuated between maximums as you drive a distance of 0.4500 km parallel to the line joining the two towers. What is the frequency of the station to which you are listening?

Homework Equations


This is supposed to test at how well we can find out data about upcoming units, so I had to guess at which equations would be most relevant since we have yet to learn them. This formula is the only one I know of that I figured would help in any way.

velocity = frequency * lambda

The Attempt at a Solution


I assumed by maximums, it means 0.45/2 on each side of the "present location", so I calculated the distance to the points from each tower.

I drew out the diagram, which made an isoceles triangle with dimensions 0.1750 km, 25 km and 25 km, then split the triangle in half to figure out the distance between the line created by the two receivers and the highway, and using Pythagorean theorem I got a value around 24.99984687 km. I then used that value to find out the distance from the maximums to the towers. From either point to the tower farthest from it, it would be 25.00179994 km away, and to the tower closest it would be 25.000225 km.

I calculated the time to reach each tower, and the values were 0.073520961 s to the closest tower and 0.073534705 s to the farthest.

I assume from here it would do something with resonance but I can not think of the correct formula to use to calculate that. I don't necessarily need a full answer, but a step in the right direction would be very helpful. Thanks in advance!

This is a classic two source interference problem:

The 450m is the 'delta X' distance between minima, the 25km is the 'L' distance from the source, the 175m is 'd' the slit separation for double slit intereference (think Young's exp).

Use the relation: 'lambda' over 'd' = 'delta X' over 'L' and solve for Lambda. Use this wavelength to solve for 'f' (frequency) keeping in mind that v = c = 3.0x10^8 m/s
 

What is the formula for calculating radio station frequency given distances?

The formula for calculating radio station frequency given distances is f = c / λ, where f is the frequency in hertz (Hz), c is the speed of light in meters per second (m/s), and λ is the wavelength in meters (m).

What is the speed of light?

The speed of light is approximately 299,792,458 meters per second (m/s). This value is used in the formula for calculating radio station frequency given distances.

How do I convert between frequency and wavelength?

To convert frequency to wavelength, use the formula λ = c / f, where λ is the wavelength in meters (m), c is the speed of light in meters per second (m/s), and f is the frequency in hertz (Hz). To convert wavelength to frequency, use the formula f = c / λ.

What units are used for frequency and wavelength?

Frequency is measured in hertz (Hz), which is the number of cycles per second. Wavelength is measured in meters (m), which is the distance between two consecutive peaks or troughs of a wave.

How accurate is the calculation of radio station frequency given distances?

The calculation of radio station frequency given distances is very accurate, as long as the speed of light is used as the constant for the calculation. Any errors in the measurement of distance or wavelength could affect the accuracy of the calculation.

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