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\(\bullet\) A wave on a rope that measures \(10 \mathrm{~m}\) long takes \(2.0 \mathrm{~s}\) to travel the whole rope. If the wavelength of the wave is \(2.5 \mathrm{~m},\) what is the frequency of oscillation of any piece of the rope?

Short Answer

Expert verified
The frequency of oscillation is 2 Hz.

Step by step solution

01

Determine the Wave Speed

To find the frequency, we first need to determine the wave speed. The wave speed \( v \) can be calculated using the formula: \[v = \frac{d}{t}\]where \( d = 10 \, \text{m} \) is the distance the wave travels (length of the rope) and \( t = 2.0 \, \text{s} \) is the time it takes to travel this distance. Thus,\[v = \frac{10 \text{ m}}{2.0 \text{ s}} = 5 \text{ m/s}.\]
02

Use the Wave Speed to Find Frequency

Having found the wave speed, we can now find the frequency of the wave. The relationship between wave speed \( v \), wavelength \( \lambda \), and frequency \( f \) is given by the formula:\[v = f \times \lambda.\]We can rearrange this to find the frequency:\[f = \frac{v}{\lambda}.\]Substitute the values \( v = 5 \text{ m/s} \) and \( \lambda = 2.5 \text{ m} \):\[f = \frac{5 \text{ m/s}}{2.5 \text{ m}} = 2 \text{ Hz}.\]

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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 crucial concept in understanding the physics of waves. It represents how fast a point on the wave, such as the crest, travels through a medium. Wave speed can be calculated using the formula: \[ v = \frac{d}{t}\]where:
  • \( v \) is the wave speed,
  • \( d \) is the distance the wave travels,
  • \( t \) is the time it takes to travel that distance.
In the context of the original exercise, the wave covers a distance of 10 meters in 2 seconds. Using our formula: \[v = \frac{10 \text{ m}}{2.0 \text{ s}} = 5 \text{ m/s}.\]This result tells us that each point on the wave travels at a speed of 5 meters per second along the rope. This understanding of wave speed is foundational for further calculations like frequency and wavelength.
Frequency Calculation
Frequency is the measure of how many wave cycles pass a point per unit of time. It's an essential characteristic of any wave and is related to how we perceive things like sound and light. Frequency is defined by the equation: \[f = \frac{v}{\lambda}\]where:
  • \( f \) is the frequency,
  • \( v \) is the wave speed,
  • \( \lambda \) is the wavelength.
By rearranging the wave speed equation, we are able to solve for frequency. For the example, with a wave speed \( v = 5 \text{ m/s} \) and a wavelength \( \lambda = 2.5 \text{ m} \), the calculation is: \[f = \frac{5 \text{ m/s}}{2.5 \text{ m}} = 2 \text{ Hz}.\]This means that the rope oscillates at a frequency of 2 Hertz, indicating 2 full wave cycles pass a point each second. Understanding frequency helps us interpret the dynamics of wave motion in various contexts.
Wavelength
Wavelength is the distance between identical points in the adjacent cycles of a wave, such as crest to crest or trough to trough. It is denoted by \( \lambda \) and plays a critical role in the dynamics of wave motion. Wavelength, along with wave speed, helps determine the frequency of the wave. It is measured in meters (m) and gives a sense of the "size" of the wave in terms of distance. In our exercise, the given wavelength is 2.5 meters, which means each full cycle of the wave covers a 2.5-meter stretch on the rope. This concept, paired with wave speed, allows us to find the frequency and provides a complete picture of a wave's nature. Having a clear understanding of wavelength is vital for solving various physics problems and provides insights into how waves propagate through different media.

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Most popular questions from this chapter

\(\bullet\bullet\) The equation of motion of a SHM oscillator is \(x=(0.50 \mathrm{~m}) \sin (2 \pi f) t,\) where \(x\) is in meters and \(t\) is in seconds. If the position of the oscillator is at \(x=0.25 \mathrm{~m}\) at \(t=0.25 \mathrm{~s},\) what is the frequency of the oscillator?

\(\bullet\bullet\) A vertical spring has a 0.200 -kg mass attached to it. The mass is released from rest and falls \(22.3 \mathrm{~cm}\) before stopping (a) Determine the spring constant. (b) Determine the speed of the mass when it has fallen only \(10.0 \mathrm{~cm}\).

\(\bullet\)Dolphins and bats determine the location of their prey using echolocation (see Conceptual Question 15\()\). If it takes \(15 \mathrm{~ms}\) for a bat to receive the ultrasonic sound wave reflected off a mosquito, how far is the mosquito from the bat? Take the speed of sound as \(345 \mathrm{~m} / \mathrm{s}\)

String A has twice the tension but half the linear mass density as string \(B\), and both strings have the same length. (a) The frequency of the first harmonic on string A is (1) four times, (2) twice, (3) half, (4) \(1 / 4\) times that of string B. Explain. (b) If the lengths of the strings are \(2.5 \mathrm{~m}\) and the wave speed on string \(\mathrm{A}\) is \(500 \mathrm{~m} / \mathrm{s},\) what are the frequencies of the first harmonic on both strings?

\(\bullet\) A sound wave has a speed of \(340 \mathrm{~m} / \mathrm{s}\) in air. If this wave produces a tone with a frequency of \(1000 \mathrm{~Hz}\), what is its wavelength?

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