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For a certain transverse wave, the distance between two successive crests is \(1.20 \mathrm{m},\) and eight crests pass a given point along the direction of travel every 12.0 \(\mathrm{s}\) . Calculate the wave speed.

Short Answer

Expert verified
The speed of the wave is 0.80 m/s.

Step by step solution

01

Determine the Wavelength

The distance between two successive crests of a transverse wave is known as the wavelength. In this case, the wavelength (\(\text{Wavelength} \text{,}\ \text{symbol} \text{,}\ \text{denoted by}\ \text{,} \text{}, \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by}\ \text{,} \text{} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by} \text{,} \text{denoted by} (\lambda\)) is given as 1.20 meters.
02

Calculate the Frequency

Frequency (\(f\)) is the number of times a specific point on the wave, such as a crest, passes a given point in a certain period of time. Here, eight crests pass in 12.0 seconds, so the frequency is the number of crests divided by the time, which calculates to \(f = \frac{8 \text{ crests}}{12.0 \text{ s}} = \frac{2}{3} \text{ Hz}\).
03

Calculate the Wave Speed

The wave speed (\(v\)) can be calculated using the formula \(v = f\cdot\lambda\), where \(\lambda\) is the wavelength and \(f\) is the frequency. Substituting the given values, we get \(v = \frac{2}{3} \text{ Hz} \cdot 1.20 \text{ m} = 0.80 \text{ m/s}\). Thus, the speed of the wave is 0.80 meters per second.

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Key Concepts

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

Wavelength
To understand wave speed, one must first be acquainted with the concept of wavelength, denoted by the Greek letter lambda ((lambda)). Wavelength is the distance between two consecutive points that are in phase on a wave, such as from crest to crest or trough to trough in transverse waves. It is a key factor in calculating the speed of a wave.

Imagine holding a rope and flicking it up and down to create waves. The wavelength would be the distance between two of the highest points (crests) that you see in the rope. In the exercise, the given wavelength of a transverse wave is measured as 1.20 meters between successive crests. This is the spatial period of the wave—the distance over which the wave's shape repeats—and it's fundamental for determining wave speed in combination with frequency.
Frequency

Understanding Frequency in Waves


Frequency is the number of occurrences of a repeating event per unit of time. In the context of waves, it refers to the number of complete cycles that pass a given point in a specified period of time, and it's measured in Hertz (Hz), where one Hertz is one cycle per second. The concept is integral to various realms of physics and everyday phenomena like sound and light.

For example, if you're watching waves roll into the shore, the frequency would be how many waves reach you within a second. Relating to the provided exercise, if eight crests pass a given point in 12.0 seconds, the frequency calculation would be 2/3 Hz, meaning there are roughly 0.67 crests passing the point every second. The frequency imparts a critical insight into the temporal aspect of the wave cycle—how often the waves occur—which, along with wavelength, will determine the wave speed.
Transverse Waves
Transverse waves are a type of wave where the motion of the medium's particles is perpendicular to the direction of the wave's travel. They are easily visualized by the up and down motion seen in waves on a string or water's surface.

Imagine you are at a stadium and observe a 'wave' of standing fans passing by—it's a visually compelling example of a transverse wave in action. Each fan stands up and sits down, moving perpendicular to the direction of the wave itself. Similarly, in our exercise, the wave on a string triggered by moving the end up and down is a transverse wave. Different from longitudinal waves, where the displacement of the medium is parallel to the wave's direction (like sound waves), transverse waves showcase phenomena like light and radio waves. Understanding these different wave types helps clarify how waves move through various media and influence the calculation of their speed.

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Most popular questions from this chapter

A sinusoidal wave on a string is described by $$ y=(0.51 \mathrm{cm}) \sin (k x-\omega t) $$ where \(k=3.10 \mathrm{rad} / \mathrm{cm}\) and \(\omega=9.30 \mathrm{rad} / \mathrm{s}\) . How far does a wave crest move in 10.0 \(\mathrm{s}\) ? Does it move in the positive or negative \(x\) direction?

Review problem. A block of mass \(M\) hangs from a rubber cord. The block is supported so that the cord is not stretched. The unstretched length of the cord is \(L_{0}\) and its mass is \(m,\) much less than \(M\) . The "spring constant" for the cord is \(k .\) The block is released and stops at the lowest point. (a) Determine the tension in the string when the block is at this lowest point. (b) What is the length of the cord in this "stretched" position? (c) Find the speed of a transverse wave in the cord if the block is held in this lowest position.

A string of length \(L\) consists of two sections. The left half has mass per unit length \(\mu=\mu_{0} / 2,\) while the right has a mass per unit length \(\mu^{\prime}=3 \mu=3 \mu_{0} / 2\) . Tension in the string is \(T_{0} .\) Notice from the data given that this string has the same total mass as a uniform string of length \(L\) and mass per unit length \(\mu_{0} .\) (a) Find the speeds \(v\) and \(v^{\prime}\) at which transverse pulses travel in the two sections. Express the speeds in terms of \(T_{0}\) and \(\mu_{0},\) and also as multiples of the speed \(v_{0}=\left(T_{0} / \mu_{0}\right)^{1 / 2} .\) (b) Find the time interval required for a pulse to travel from one end of the string to the other. Give your result as a multiple of \(\Delta t_{0}=L / v_{0} .\)

It is stated in Problem 59 that a pulse travels from the bottom to the top of a hanging rope of length \(L\) in a time interval \(\Delta t=2 \sqrt{L / g}\) . Use this result to answer the following questions. (It is not necessary to set up any new integrations.) (a) How long does it take for a pulse to travel halfway up the rope? Give your answer as a fraction of the quantity 2\(\sqrt{L / g}\) . (b) A pulse starts traveling up the rope. How far has it traveled after a time interval \(\sqrt{L / g ?}\)

S and P waves, simultaneously radiated from the hypocenter of an earthquake, are received at a seismographic station 17.3 \(\mathrm{s}\) apart. Assume the waves have traveled over the same path at speeds of 4.50 \(\mathrm{km} / \mathrm{s}\) and 7.80 \(\mathrm{km} / \mathrm{s}\) . Find the distance from the seismograph to the hypocenter of the quake.

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