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Chapter 16: Problem 14

Find the wavelength of a wave traveling at \(2.68 \times 10^{6} \mathrm{~m} / \mathrm{s}\) with a period of \(0.0125 \mathrm{~s} .\)

Short Answer

Expert verified
The wavelength is \(3.35 \times 10^{4} \; \text{m}\).

Step by step solution

01

Write out known values

First, identify the given information in the exercise. The speed of the wave is given as \(v = 2.68 \times 10^{6} \; \text{m/s}\) and the period is \(T = 0.0125\; \text{s}\).
02

Recall the relationship between speed, wavelength, and period

The relationship between the speed \(v\), wavelength \(\lambda\), and period \(T\) of a wave is given by the formula: \[ v = \lambda \times f \]where \(f\) is the frequency. Since \(f = \frac{1}{T}\), substituting for \(f\) gives us:\[ v = \lambda \times \frac{1}{T} \] or equivalently \[ \lambda = v \times T \].
03

Calculate wavelength using the formula

Substitute the known values into the wavelength formula to find \(\lambda\):\[ \lambda = (2.68 \times 10^{6} \; \text{m/s}) \times 0.0125 \; \text{s} \].Perform the multiplication:\[ \lambda = 3.35 \times 10^{4} \; \text{m} \].
04

Write the final result

Conclude that the wavelength of the wave is \(3.35 \times 10^{4} \; \text{m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Speed
Wave speed is a key concept when studying how waves travel through different mediums. It tells us how fast a wave propagates through a medium and is usually measured in meters per second (m/s).
To calculate wave speed, you need to consider two aspects: the frequency of the wave and its wavelength. The basic formula connecting these is:
  • \[ v = ext{wavelength} ( \lambda) imes ext{frequency} (f) \]
The speed of a wave depends on the type of wave and the medium it's moving through. For example, sound waves travel at different speeds in air, water, and solids.
In our exercise, the wave speed is given as \(2.68 imes 10^{6} ext{ m/s}\). This information will help us find other properties of the wave, such as its wavelength.
Wave Period
The wave period, denoted as \(T\), is the time it takes for a single cycle of the wave to pass a given point. It is often measured in seconds.
The period is closely related to frequency since frequency represents how many waves pass through a point in one second. The two are inversely related:
  • The formula to express this relationship is: \[ f = \frac{1}{T} \]
If you know one, you can easily find the other.
In our case, the period is given as \(0.0125 ext{ s}\). This will help us calculate the frequency, which is then used to find the wavelength.
Frequency Calculation
Calculating the frequency is crucial for solving wave-related physics problems. Frequency, represented by \(f\), tells us how many wave cycles pass through a point in one second. It's measured in Hertz (Hz).
We can calculate frequency if we know the period using the formula:
  • \[ f = \frac{1}{T} \]
This relationship helps connect frequency and period directly. With a period of \(0.0125 ext{ s}\), the frequency \(f\) comes out to be \(\frac{1}{0.0125} \approx 80 ext{ Hz}\).
Knowing frequency allows us to explore and understand more characteristics of the wave, like determining its wavelength using the speed of the wave and the formula \( v = \lambda \times f \).
Physics Problem Solving
Solving physics problems often requires understanding a few key principles related to the concepts involved. For wave problems, it is often about understanding the interplay between wave speed, period, and frequency.
Here鈥檚 a step-by-step strategy for solving these problems:
  • Identify what is known:
    • Extract given values like speed, period, or frequency from the problem.
  • Understand the relationships:
    • Recall the core formulas such as \(v = \lambda \times f\) and \(f = \frac{1}{T}\).
  • Substitute values:
    • Insert the known values into these equations to find unknowns.
  • Calculate and conclude:
    • Perform calculations step-by-step to arrive at the solution.
For our specific exercise, solving complex equations becomes manageable once we substitute the known parameters鈥攚ave speed and period鈥攊nto the necessary formulas, stepping through each calculation to find the wavelength.

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Most popular questions from this chapter

A sonar receiver detects a reflected sound wave from another ship \(3.52 \mathrm{~s}\) after the wave was transmitted. How far away is the other ship? Assume that the water temperature is \(25^{\circ} \mathrm{C}\).

What is the period of a pendulum that is \(0.25 \mathrm{~m}\) long?

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

A crop duster airplane flies overhead at \(44.7 \mathrm{~m} / \mathrm{s}\). The frequency of the sound is 605 Hz. (a) What do you perceive as the frequency as it approaches? Assume the velocity of sound in air is \(345 \mathrm{~m} / \mathrm{s}\). (b) What frequency do you hear as the plane flies away and has accelerated to \(55.0 \mathrm{~m} / \mathrm{s}\) ?

A grandfather clock has a \(0.750\) -m pendulum. What is its period?
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