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Chapter 16: Problem 10

A jetskier is moving at 8.4 m/s in the direction in which the waves on a lake are moving. Each time he passes over a crest, he feels a bump. The bumping frequency is 1.2 Hz, and the crests are separated by 5.8 m. What is the wave speed?

Short Answer

Expert verified
The wave speed is approximately 10.08 m/s.

Step by step solution

01

Understand the Given Information

The jetskier's speed is 8.4 m/s, the bumping frequency is 1.2 Hz, and the distance between successive wave crests (wavelength, \( \lambda \)) is 5.8 m. We are required to find the wave speed (\( v \)).
02

Calculate Relative Wave Frequency

The frequency of bumps (1.2 Hz) is the frequency with which the jetskier encounters wave crests relative to their own speed. This is a combination of the frequency of the actual waves and the jetskier's speed.
03

Use the Wave Frequency Mathematical Relationship

The apparent frequency \( f_{apparent} \) is given by \( f_{apparent} = f_{wave} + \frac{v_{jetskier}}{\lambda} \), where \( f_{wave} \) is the wave frequency, which we need to find.
04

Rearrange the Equation

Rearrange to solve for the wave frequency \( f_{wave} \): \[ f_{wave} = f_{apparent} - \frac{v_{jetskier}}{\lambda} \]
05

Calculate the Wave Frequency

Substitute the known values into the equation:\[ f_{wave} = 1.2 \, \text{Hz} - \frac{8.4 \, \text{m/s}}{5.8 \, \text{m}} = 1.2 \, \text{Hz} - 1.448 \, \text{Hz} \approx -0.248 \, \text{Hz} \].However, since frequency cannot be negative, this indicates an error in presentation. This calculation reveals \( v_{wave} > v_{jetskier} \).
06

Consider Reverse Calculation for Wave Speed

The apparent frequency indicates \( v_{wave} \) must surpass \( v_{jetskier} \). Calculate the wave speed correctly:\[ v_{wave} = \frac{1.2 \, \text{Hz} \times 5.8 \, \text{m} + 8.4 \, \text{m/s}}{1} = \frac{15.12 \, \text{m/s} + 8.4 \, \text{m/s}} \approx 10.08 \, \text{m/s} \]
07

Conclude with Corrected Wave Speed

The error in the previous step due to negative frequency highlights that misinterpretation corrected with logical reasoning and real-world constraints indicates wave speed positively: Therefore,\[ v_{wave} = 10.08 \, \text{m/s} \] is swiftly reconfirmed avoiding issues in negative presentation.

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Key Concepts

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

Wave Frequency
Wave frequency refers to the number of wave crests that pass a specific point per second. It's essentially how often a wave cycle repeats over time. In our jetski scenario, the passage over each wave crest results in a 'bump,' giving us a measure of the bumping frequency at 1.2 Hz.
This bumping frequency is crucial because it informs us of the jetskier's interaction with the waves. In simpler terms:
  • Frequency ( ext ext{Hz}), ext ext{represents how many bumps or waves the jetskier encounters per second.
  • Determines the speed of the waves relative to the observer (jetski).
To find the actual wave frequency, consider the jetskier's speed along with the wavelength, since the relative motion affects this apparent frequency.
Wavelength
Wavelength is the distance between two successive wave crests. A key term here is "successive," meaning one right after the other without skipping. The exercise specifies this distance as 5.8 meters.
Wavelength helps determine both the speed and frequency of waves. Let's break it down:
  • A longer wavelength means the crests are farther apart, affecting the frequency and speed.
  • Wavelength directly affects the wave's ability to cover distances within a certain time.
Knowing the wavelength allows us to calculate the wave speed when combined with frequency information.
Apparent Frequency
Apparent frequency describes the observed frequency when there's relative motion between the source of the wave and the observer. It's a bit like how the pitch of a siren changes as an ambulance passes by you.
In our jetski example, the jetskier moving at 8.4 m/s against the wave changes how frequently he encounters wave crests, hence affecting the apparent frequency.
Key points about apparent frequency:
  • It combines the actual wave frequency and the additional frequency induced by the observer's motion.
  • Helps explain why the jetskier experiences a different frequency than if they were stationary.
This perceived frequency can be computed using: \(f_{apparent} = f_{wave} + \frac{v_{jetskier}}{\lambda}\), integrating both the object's motion and the wave's inherent frequency.
Jetski Motion Analysis
Analyzing the motion of a jetskier involves understanding how their speed affects the perception of wave properties such as frequency and speed. When a jetskier moves at 8.4 m/s, the motion contributes to the change in apparent frequency, which can be calculated through the formula stated earlier.
For a complete motion analysis, consider:
  • The jetskier's direction and speed in relation to wave direction.
  • How this motion changes the interaction with wave crests, thus affecting perceived wave bumping frequency.
If you're trying to find wave speed, you will employ jetski speed, apparent frequency, and wavelength information 鈥� putting it all together for a coherent picture. The calculated wave speed, as per the corrected final step, is approximately 10.08 m/s, confirming how the jetski's motion intertwines with wave behavior.

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Most popular questions from this chapter

(a) A uniform rope of mass m and length L is hanging straight down from the ceiling. A small-amplitude transverse wave is sent up the rope from the bottom end. Derive an expression that gives the speed v of the wave on the rope in terms of the distance y above the bottom end of the rope and the magnitude g of the acceleration due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of 0.50 m and 2.0 m above the bottom end of the rope.

Light is an electromagnetic wave and travels at a speed of \(3.00 \times 10^{8} \mathrm{m} / \mathrm{s}\) . The human eye is most sensitive to yellow-green light, which has a wavelength of \(5.45 \times 10^{-7} \mathrm{m} .\) What is the frequency of this light?

A copper wire, whose cross-sectional area is \(1.1 \times 10^{-6} \mathrm{m}^{2},\) has a linear density of \(9.8 \times 10^{-3} \mathrm{kg} / \mathrm{m}\) and is strung between two walls, At the ambient temperature, a transverse wave travels with a speed of 46 \(\mathrm{m} / \mathrm{s}\) on this wire. The coefficient of linear expansion for copper is \(17 \times 10^{-6}\left(\mathrm{C}^{9}\right)^{-1},\) and Young's modulus for copper is \(1.1 \times 10^{11} \mathrm{N} / \mathrm{N}^{2}\) . What will be the speed of the wave when the temperature is lowered by 14 \(\mathrm{C}^{\circ}\) ? Ignore any change in the linear density caused by the change in temperature.

An explosion occurs at the end of a pier. The sound reaches the other end of the pier by traveling through three media: air, fresh water, and a slender metal handrail. The speeds of sound in air, water, and the handrail are 343, 1482, and 5040 m/s, respectively. The sound travels a distance of 125 m in each medium. (a) Through which medium does the sound arrive first, second, and third? (b) After the first sound arrives, how much later do the second and third sounds arrive?

The middle C string on a piano is under a tension of 944 N. The period and wavelength of a wave on this string are 3.82 ms and 1.26 m, respectively. Find the linear density of the string.
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