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Chapter 16: Problem 45

A swimming duck paddles the water with its feet once cvery 1.6 \(\mathrm{s}\) , producing surface waves with this period. The duck is moving at constant speed in a pond where the speed of surface waves is 0.32 \(\mathrm{m} / \mathrm{s}\) , and the crests of the waves ahead of the duck are spaced 0.12 \(\mathrm{m}\) apart. (a) What is the duck's speed? (b) How far apart are the crests behind the duck?

Short Answer

Expert verified
(a) Duck's speed is 0.245 m/s. (b) Crests behind the duck are 0.12 m apart.

Step by step solution

01

Understand the Concept of Wave Speed

The speed of a wave, \( v \), is related to its frequency \( f \) and wavelength \( \lambda \) via the equation: \( v = f \cdot \lambda \). Here, the frequency is the inverse of the period, \( f = \frac{1}{T} \). The duck affects the wavelength of the waves in front of and behind it because it moves relative to the medium.
02

Determine Frequency of the Waves

Given the period \( T = 1.6 \) s, calculate the frequency \( f \) using \( f = \frac{1}{T} \). This yields \( f = \frac{1}{1.6} = 0.625 \) Hz.
03

Calculate the Duck's Speed

Calculate the speed of the duck \( v_d \). It moves towards the wave fronts, so the observed wave speed in front of the duck is the wave speed plus the duck's speed. Let \( \lambda_{ahead} = 0.12 \) m be the wavelength ahead of the duck. Then, \( 0.32 \) m/s (speed of waves) = \( v_d + 0.625 \times \lambda_{ahead} = v_d + 0.625 \times 0.12 \). Solve for \( v_d \).
04

Solve for the Duck's Speed

Substituting the known values in the equation from Step 3: \( 0.32 = v_d + 0.075 \) gives \( v_d = 0.32 - 0.075 = 0.245 \) m/s.
05

Calculate Wavelength Behind the Duck

For crests behind the duck, the effective wave speed is the actual wave speed minus the duck's speed. \( v = 0.32 - 0.245 = 0.075 \). Use the wave equation with this new speed: \( 0.075 = 0.625 \times \lambda_{behind} \). Solve for \( \lambda_{behind} \).
06

Solve for Wavelength Behind the Duck

Rearranging the equation \( 0.075 = 0.625 \times \lambda_{behind} \), solve for \( \lambda_{behind} \). \( \lambda_{behind} = \frac{0.075}{0.625} = 0.12 \) m, which surprisingly remains the same.

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Key Concepts

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

Frequency
To better understand wave motion, let's dive into the concept of frequency. Frequency refers to how often the crests of a wave pass a specific point in a certain amount of time. It's typically measured in hertz (Hz), where 1 Hz equals one wave cycle per second.
This is closely connected to the wave's period, which is the time it takes for one wave cycle to pass. The relationship is simple: frequency is the inverse of the period. So, if a wave has a period of 1.6 seconds, the frequency would be \(f = \frac{1}{1.6} = 0.625 \text{ Hz}\). This indicates the wave passes through 0.625 cycles each second.
Wavelength
Wavelength is the distance between two consecutive crests (or troughs) of a wave. It's an essential factor in determining wave speed. In our scenario with the duck, the initial wavelength ahead of the duck is 0.12 meters.
The effective wavelength can change based on the motion of the source of the waves鈥攈ere, the duck moving through water. But note how the calculated wavelength behind the duck remained at 0.12 meters! This physically means that the duck's motion didn't compress or stretch the waves trailing behind it.
Doppler Effect
The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer moving relative to the source of the waves. This effect is noticeable when the duck moves through the pond. It's similar to hearing a siren from a moving ambulance, where the pitch changes based on the vehicle's position relative to you.
In terms of our duck example, as the duck advances, the wave crests in front of it are compressed, resulting in reduced wavelength (though in this problem, both wavelengths appear equal, which can sometimes happen). The waves behind the duck might get stretched, exhibiting a potential increase in wavelength, thus illustrating the Doppler effect.
Wave Equation
The wave equation is fundamental in connecting the properties of waves: speed, frequency, and wavelength. It is given as \( v = f \cdot \lambda \), where \(v\) is the wave speed, \(f\) the frequency, and \(\lambda\) the wavelength.
In our exercise, we use this equation to determine how the duck's movement alters perceived wave speeds. For example, the wave speed ahead of the duck is the sum of the water wave speed and the duck's speed, showing how motion adds to wave perception. Similarly, wave speed behind the duck subtracts the duck's speed, altering the resulting wavelength calculation.

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Most popular questions from this chapter

The intensity due to a number of independent sound sources is the sum of the individual intensities. (a) When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries? (b) To increase the sound intensity level again by the same number of decibels as in part (a), how many more crying babies are required?

Longitudinal Waves in Different Fluids. (a) A longitudinal wave propagating in a water-filled pipe has intensity \(3.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) and frequency 3400 \(\mathrm{Hz}\) . Find the amplitude \(A\) and wavelength \(\lambda\) of the wave. Water has density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) and bulk modulus \(2.18 \times 10^{9} \mathrm{Pa}\) . Water has density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) at pressure \(1.00 \times 10^{5} \mathrm{Pa}\) and density 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) , what will be the amplitude \(A\) and wavelength \(\lambda\) of a longitudinal wave with the same intensity and frequency as in part (a)? (c) In which fluid is the amplitude larger, water or air? What is the ratio of the two amplitudes? Why is this ratio so different from 1.00\(?\)

A railroad train is traveling at 30.0 \(\mathrm{m} / \mathrm{s}\) in still air. The frequency of the note cmitted by the train whistle is 262 \(\mathrm{Hz}\) . What frequency is heard by a passenger on a train moving in the opposite direction to the first at 18.0 \(\mathrm{m} / \mathrm{s}\) and (a) approaching the first; and (b) receding from the first?

A soprano and a bass are singing a duet. While the soprano \(\operatorname{sings}\) an \(A^{\prime \prime}\) at 932 . Hz, the bass sings an \(A^{\prime \prime}\) but three octaves lower. In this concert hall, the density of air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its bulk modulus is \(1.42 \times 10^{5} \mathrm{Pa}\) . In order for their notes to have the same Rnund intensity level, what must he (a) the ratio of the pressur amplitude of the bass to that of the soprano, and (b) the ratio of the displacement amplitude of the bass to that of the soprano? (c) What displacement amplitude (in \(\mathrm{m}\) and \(\mathrm{nm}\) ) does the soprano produce to sing her \(A^{\prime \prime}\) at 72.0 \(\mathrm{dB} ?\)

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we recelve from it is 10.0\(\%\) higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)
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