Problem 15 - Applied Physics

Question

Find the wavelength of a wave traveling twice the speed of sound (speed of sound \(=331 \mathrm{~m} / \mathrm{s}\) ) that is produced by an oscillator emitting 63 pulses every \(8.3 \times 10^{-6} \mathrm{~min}\).

Step-by-Step Solution

Answer

The wavelength of the wave is approximately 0.0523 meters.

Step 1: Convert Time to Seconds

First, convert the time given in minutes to seconds. We know that there are 60 seconds in a minute, so the time in seconds is: \[ 8.3 \times 10^{-6} \text{ min} \times 60 \text{ s/min} = 4.98 \times 10^{-4} \text{ s} \]

Step 2: Calculate the Frequency

Next, find the frequency of the oscillator. The frequency \( f \) is the number of pulses divided by the time in seconds: \[ f = \frac{63}{4.98 \times 10^{-4}} \approx 12650.6 \text{ Hz} \]

Step 3: Find the Speed of the Wave

The speed of the wave is twice the speed of sound. Given the speed of sound is 331 m/s, the speed \( v \) of the wave is: \[ v = 2 \times 331 \text{ m/s} = 662 \text{ m/s} \]

Step 4: Calculate the Wavelength

Finally, calculate the wavelength \( \lambda \) using the formula \( v = f \times \lambda \). Rearrange the formula to solve for wavelength: \[ \lambda = \frac{v}{f} = \frac{662}{12650.6} \approx 0.0523 \text{ m} \]

Key Concepts

Wavelength Calculation

In the study of waves, understanding how to calculate the wavelength is very important. The wavelength, often represented by the Greek letter \( \lambda \), is the distance between two consecutive points of a wave that are in phase, such as two crests or troughs. To calculate the wavelength, you can use the formula:
\[ \lambda = \frac{v}{f} \]
Here, \( v \) is the speed of the wave and \( f \) is its frequency. In our example, we know the wave travels faster than the speed of sound. Specifically, it moves at twice this speed, or \( 662 \text{ m/s} \). We've also calculated the frequency as \( 12650.6 \text{ Hz} \).

• Perform the division \( \lambda = \frac{662}{12650.6} \).
• This results in \( \lambda \approx 0.0523 \text{ meters} \).

With these calculations, you can understand the process of determining the wavelength of any wave, given the speed and frequency.

Frequency of Waves

The frequency of a wave refers to how often the wave cycles occur each second. It is a critical measure and is typically denoted by the symbol \( f \), with units in Hertz (Hz).
In the given exercise, the frequency of the oscillator is determined by the formula:
\[ f = \frac{\text{number of pulses}}{\text{time in seconds}} \]
From the problem, we know there are 63 pulses occurring within \( 4.98 \times 10^{-4} \) seconds. Calculating gives us:
\[ f = \frac{63}{4.98 \times 10^{-4}} \approx 12650.6 \text{ Hz} \]

• Frequency is a measure of cycles per second.
• In our scenario, \( 12650.6 \text{ cycles per second} \).

Recognizing the frequency provides insight into the energy and type of wave we are dealing with.

Speed of Sound

The speed of sound is a fundamental concept in wave physics. It is the speed at which sound waves travel through a medium, like air, and is influenced by factors such as temperature and pressure.
In most exercises, a typical reference speed is given. For example:

• Assumed speed at room temperature: \( 331 \text{ m/s} \).

For this problem, knowing the wave travels twice this speed is crucial. Therefore:\[ 2 \times 331 \text{ m/s} = 662 \text{ m/s} \] is the speed of our wave.

• Wind, weather, or medium changes can alter sound speed.
• Sound travels faster in warmer environments.

Understanding the speed of sound is essential when dealing with problems involving sound waves and their properties.