Problem 6 - Applied Physics

Question

Find the wavelength of water waves with frequency \(0.650 \mathrm{~Hz}\) and velocity \(1.50 \mathrm{~m} / \mathrm{s}\).

Step-by-Step Solution

Answer

The wavelength is approximately 2.31 meters.

Step 1: Understanding the Problem

We need to find the wavelength of water waves. We are given the frequency, \(f = 0.650 \, \mathrm{Hz}\), and the velocity of the waves, \(v = 1.50 \, \mathrm{m/s}\). The relationship between velocity, frequency, and wavelength is given by the formula \(v = f \cdot \lambda\), where \(\lambda\) is the wavelength.

Step 2: Rearranging the Formula

To find the wavelength \(\lambda\), we rearrange the formula \(v = f \cdot \lambda\) to solve for \(\lambda\). This gives us \(\lambda = \frac{v}{f}\).

Step 3: Substitute the Values

We substitute the given values for velocity and frequency into the rearranged equation. So, \(\lambda = \frac{1.50}{0.650}\).

Step 4: Calculate the Wavelength

Perform the division to find the wavelength. \(\lambda = \frac{1.50}{0.650} \approx 2.31 \text{ meters}\).

Key Concepts

Wavelength Calculation

When discussing wave phenomena, one essential characteristic is the wavelength. The wavelength is the distance between successive crests or troughs of a wave. It's usually denoted by the Greek letter lambda, \(\lambda\). To find the wavelength when you know the frequency (\(f\)) and velocity (\(v\)) of a wave, you can use the equation:

• \(v = f \cdot \lambda\)
• Rearranged, this becomes \(\lambda = \frac{v}{f}\)

Let's consider the problem presented: you need to find the wavelength of water waves with a frequency of \(0.650\, \text{Hz}\) and a velocity of \(1.50\, \text{m/s}\). Rearranging the formula to \(\lambda = \frac{1.50}{0.650}\) and solving gives us a wavelength of approximately \(2.31\, \text{meters}\).
This method of calculation can be applied to any wave, whether it's water, sound, or light, as long as the frequency and velocity are known.

Wave Frequency

Frequency is a fundamental property of waves. It refers to how often the waves occur over a specific time period. In most physics problems, the frequency is given in Hertz (Hz), which equates to cycles (or waves) per second. Understanding wave frequency is crucial because it ties directly to the wave's energy and behavior.
In our example, the frequency given is \(0.650\, \text{Hz}\). This tells us that the waves pass a given point at 0.650 cycles every second. Frequency directly influences the wavelength and velocity, as seen in the wave speed equation \(v = f \cdot \lambda\).

• Higher frequency means shorter wavelength if velocity is constant.
• Lower frequency results in a longer wavelength.

This relationship helps explain why sounds of different frequencies are heard differently and why certain light frequencies give us certain colors.

Wave Velocity

The velocity of a wave refers to how fast the wave crest moves through space. It is typically measured in meters per second (m/s). In wave physics, velocity is essential for determining how quickly a wave propagates through a given medium. Different waves have varying velocities depending on the medium and the type of wave.
In our initial problem, the velocity of the water waves is \(1.50\, \text{m/s}\). This indicates that each crest of the water wave moves 1.50 meters every second. The velocity of a wave can be measured directly or calculated using the frequency and wavelength with the equation:

• \(v = f \cdot \lambda\)

In constant conditions, if the frequency increases, yet the velocity stays constant, the wavelength must decrease. Conversely, if the frequency decreases and velocity remains constant, the wavelength must increase.
Understanding this concept helps explain natural phenomena like the speed of sound in air or the speed of light in a vacuum.