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Chapter 22: Problem 25

A transverse wave is set up on a taut string. Its free end is wiggled up and down at a rate of \(10.0\) cycles every second. What happens to the wavelength of the waves when the oscillation rate is raised to \(20.0\) cycles per second, all else kept constant?

Short Answer

Expert verified
The wavelength is halved when the frequency is doubled.

Step by step solution

01

Recall the Wave Speed Formula

The wave speed is given by the formula: \[ v = f \lambda \]where \( v \) is the wave speed, \( f \) is the frequency, and \( \lambda \) is the wavelength. Since we're told that all else is constant, the wave speed \( v \) does not change.
02

Calculate Initial Frequency and Wavelength Relationship

Initially, the frequency \( f_1 \) is 10 cycles per second. Let's denote its corresponding wavelength as \( \lambda_1 \). From our formula:\[ v = f_1 \lambda_1 = 10 \lambda_1 \]
03

Apply New Frequency and Calculate New Wavelength

When the frequency \( f_2 \) is changed to 20 cycles per second, we have:\[ v = f_2 \lambda_2 = 20 \lambda_2 \]Since the wave speed \( v \) does not change, we can set the two expressions for \( v \) equal to each other:\[ 10 \lambda_1 = 20 \lambda_2 \]
04

Solve for the New Wavelength

Re-arrange the equation to find \( \lambda_2 \):\[ \lambda_2 = \frac{10}{20} \lambda_1 = \frac{1}{2} \lambda_1 \] This shows that the new wavelength \( \lambda_2 \) is half of the original wavelength \( \lambda_1 \).

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Key Concepts

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

Frequency
In wave mechanics, frequency refers to how often waves occur in a given unit of time. It's measured in cycles per second, or Hertz (Hz). In the context of our problem, the frequency tells us how many peaks of the wave pass a fixed point in one second. For example, an increase from 10 cycles per second to 20 cycles per second effectively doubles the frequency.

Frequency is a crucial concept because it dictates wave behavior when other conditions are constant. It influences the pitch of sounds and the color of light. In our specific exercise, as the rate of wave oscillation increased (from 10 Hz to 20 Hz), this frequency change directly impacted the wavelength. Understanding how frequency interacts with other properties of a wave like wavelength and wave speed is fundamental for comprehending wave mechanics.
Wavelength
Wavelength is the distance between consecutive peaks or troughs in a wave. It plays a vital role in defining the characteristics of a wave. In our exercise, we explored how changing the frequency affects the wavelength when the wave speed remains the same.

The wavelength is inversely related to frequency, which means as frequency increases, wavelength decreases and vice versa. In the formula \[ v = f \lambda \] where \( v \) is the wave speed, \( f \) is the frequency, and \( \lambda \) is the wavelength. As shown in the problem, doubling the frequency from 10 cycles per second to 20 cycles per second results in the wavelength becoming half, demonstrating this inverse relationship clearly. Because wave speed \( v \) remains constant, any change in frequency needs an opposite change in wavelength.
Wave Speed
Wave speed is a measure of how fast a wave propagates through a medium. It is an essential concept for understanding wave dynamics. The formula for wave speed is \( v = f \lambda \) showing that it is the product of frequency \( f \) and wavelength \( \lambda \).

In our exercise, the wave speed was kept constant to isolate how changes in frequency and wavelength relate to one another. This is a common scenario in practical problems and helps emphasize the relationship between frequency and wavelength in the same medium. Keeping wave speed constant means that any increase in frequency leads to a decrease in wavelength and vice versa. Grasping this will help in solving complex problems where wave speed cannot change because the medium is unchanged.
Transverse Waves
Transverse waves are waves in which the displacement of the medium is perpendicular to the direction of wave travel. These are the kinds of waves we typically observe in strings or on the surface of liquids. In transverse waves, we can easily observe crests (the highest points) and troughs (the lowest points).

Our problem focused on transverse waves on a taut string, emphasizing the relationship between frequency, wavelength, and wave speed, while keeping wave motion perpendicular to the string. Understanding transverse waves involves recognizing how changes in the medium or frequency affect the wave's appearance and behavior. Transverse waves are crucial in many scientific applications, including understanding electromagnetic radiation, as the displacement direction provides clear insight into wave patterns and energy transmission.

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Most popular questions from this chapter

A uniform flexible cable is \(20 \mathrm{~m}\) long and has a mass of \(5.0\) kg. It hangs vertically under its own weight and is vibrated (perpendicularly) from its upper end with a frequency of \(7.0 \mathrm{~Hz}\). (a) Find the speed of a transverse wave on the cable at its midpoint. (b) What are the frequency and wavelength at the midpoint? (b) Because wave crests do not pile up along a string or cable, the number passing one point must be the same as that for any other point. Therefore, the frequency, \(7.0 \mathrm{~Hz}\), is the same at all points. To find the wavelength at the midpoint, we must use the speed we found for that point, \(9.9 \mathrm{~m} / \mathrm{s}\). That gives us $$ \lambda=\frac{v}{f}=\frac{9.9 \mathrm{~m} / \mathrm{s}}{7.0 \mathrm{~Hz}}=1.4 \mathrm{~m} $$

A string has both its total mass and length doubled; all else kept constant, what happens to the speed of transverse waves that can be set up on the string?

A metal bar \(6.0 \mathrm{~m}\) long, clamped at its center and vibrating longitudinally in such a manner that it gives its first overtone, vibrates in unison with a tuning fork marked 1200 vibration/s. Compute the speed of sound in the metal.

A string \(2.0 \mathrm{~m}\) long is driven by a 240 -Hz vibrator at its end. The string resonates in four segments forming a standing wave pattern. What would be the speed of a transverse wave on such a string? Let's first determine the wavelength of the wave from part (d) of Since each segment is \(\lambda / 2\) long, $$ 4\left(\frac{\lambda}{2}\right)=L \quad \text { or } \quad \lambda=\frac{L}{2}=\frac{2.0 \mathrm{~m}}{2}=1.0 \mathrm{~m} $$ Then, using \(\lambda=u T=v / f\), $$ \mathrm{u}=f \lambda=\left(240 \mathrm{~s}^{-1}\right)(1.0 \mathrm{~m})=0.24 \mathrm{~km} / \mathrm{s} $$

A glass tube that is \(70.0 \mathrm{~cm}\) long is open at both ends. Find the frequencies at which it will resonate when driven by sound waves that have a speed of \(340 \mathrm{~m} / \mathrm{s}\). A pipe that is open at both ends must have an antinode at each end. It will therefore resonate as in From the diagram it can be seen that the resonance wavelengths \(\lambda_{n}\) are given by $$ L=n\left(\frac{\lambda_{n}}{2}\right) \quad \text { or } \quad \lambda_{n}=\frac{2 L}{n} $$ where \(n\) is an integer. But \(\lambda_{n}=v / f_{n}\), therefore $$ f_{n}=\left(\frac{n}{2 L}\right)(v)=(n)\left(\frac{340 \mathrm{~m} / \mathrm{s}}{2 \times 0.700 \mathrm{~m}}\right)=243 n \mathrm{~Hz} $$
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