Problem 17 - Applied Physics

Question

Find the velocity of microwaves having wavelength \(0.750 \mathrm{~m}\) and frequency \(2.75 \times 10^{10} \mathrm{~Hz}\)

Step-by-Step Solution

Answer

The velocity of the microwaves is \(2.0625 \times 10^{10} \mathrm{~m/s}\).

Step 1: Understanding the Wave Equation

The velocity of a wave can be determined using the wave equation: \[ v = \lambda \cdot f \]where \(v\) is the velocity, \(\lambda\) is the wavelength, and \(f\) is the frequency.

Step 2: Identifying Given Values

From the problem, the given wavelength \(\lambda\) is \(0.750 \mathrm{~m}\) and the frequency \(f\) is \(2.75 \times 10^{10} \mathrm{~Hz}\).

Step 3: Substituting Values into the Equation

Substitute the given values into the wave equation:\[ v = 0.750 \times 2.75 \times 10^{10} \]

Step 4: Performing the Calculation

Calculate the velocity by performing the multiplication:\[ v = 0.750 \times 2.75 \times 10^{10} = 2.0625 \times 10^{10} \]

Step 5: Finalize the Result

The velocity of the microwaves is \[ v = 2.0625 \times 10^{10} \mathrm{~m/s} \]

Key Concepts

Microwave Velocity

The term "microwave velocity" refers to the speed at which microwave signals travel through a medium, usually air or vacuum. Velocity is a fundamental property of waves, determining how quickly energy is transferred from one point to another. For microwaves, as with all electromagnetic waves, this velocity can be calculated using the wave equation, which relates the velocity to both the wavelength and the frequency of the wave. Understanding microwave velocity is essential in fields such as telecommunications and radar technology, where precise knowledge of wave travel is crucial for accurate signal transmission and reception.
To calculate the velocity of microwaves, we use the formula:

• \( v = \lambda \cdot f \) where \( v \) represents velocity.
• \( \lambda \) stands for wavelength.
• \( f \) indicates frequency.

This relationship shows that the velocity depends directly on both wavelength and frequency—two key characteristics of waves.

Wavelength and Frequency

Wavelength and frequency are critical aspects of understanding microwaves and wave behavior in general. The wavelength \( \lambda \) is the distance between two consecutive peaks (or troughs) in a wave and is typically measured in meters. In our problem, the wavelength is given as \(0.750 \mathrm{~m}\).
Frequency \( f \) describes how many wave cycles pass a fixed point per unit of time, measured in hertz (Hz). In this case, the frequency is \( 2.75 \times 10^{10} \mathrm{~Hz} \).
The wave equation \( v = \lambda \cdot f \) neatly ties these two parameters with velocity. As frequency increases—indicating more waves are passing per second—velocity will increase if wavelength remains constant.
When working with electromagnetic waves like microwaves, realizing that the speed of light in a vacuum (\( c \approx 3 \times 10^8 \mathrm{~m/s} \)) is the upper limit of velocity helps place bounds on wave calculations and is crucial in practical applications.

Wave Calculation Steps

Calculating microwave velocity involves following a series of logical steps that rely on understanding and utilizing the wave equation. Here's how you solve such a problem using a step-by-step approach:

1. **Understand the Wave Equation:** Begin with the base equation \( v = \lambda \cdot f \). This provides the framework for the calculation, showing the relationship between velocity, wavelength, and frequency. 2. **Identify Given Values:** Extract information from the problem. Here, \( \lambda = 0.750 \mathrm{~m} \) and \( f = 2.75 \times 10^{10} \mathrm{~Hz} \). 3. **Substitute Values:** Plug these known values into the wave equation: \( v = 0.750 \times 2.75 \times 10^{10} \). 4. **Calculate Velocity:** Perform the arithmetic to find \( v \). Multiplying these figures gives \( v = 2.0625 \times 10^{10} \mathrm{~m/s} \). 5. **Review the Result:** The calculated velocity reflects how fast the microwaves travel in the given context, matching expectations for electromagnetic wave speeds in air or vacuum. By following these methods carefully, anyone can compute the necessary wave characteristics confidently and accurately for various practical settings.