Problem 55 - University Physics with Modern Physics

Question

CALC Ant Joy Ride. You place your pet ant Klyde (mass \(m )\) on top of a horizontal, stretched rope, where he holds on tightly. The rope has mass \(M\) and length \(L\) and is under tension \(F .\) You start a sinusoidal transverse wave of wavelength \(\lambda\) and amplitude \(A\) propagating along the rope. The motion of the rope is in a vertical plane. Klyde's mass is so small that his presence has no effect on the propagation of the wave. (a) What is Klyde's top speed as he oscillates up and down? (b) Klyde enjoys the ride and begs for more. You decide to double his top speed by changing the tension while keeping the wavelength and amplitude the same. Should the tension be increased or decreased, and by what factor?

Step-by-Step Solution

Answer

(a) Top speed is \( \frac{2\pi A}{\lambda} \sqrt{\frac{F}{\mu}} \). (b) Increase tension by a factor of 4.

Step 1: Understand the Problem

We are given an ant Klyde sitting on a rope, which has a sinusoidal wave propagating through it. We need to calculate Klyde's top speed as he oscillates and then determine how to change the tension to double this speed while maintaining the same wavelength and amplitude.

Step 2: Use the Wave Speed Formula

The speed of a wave on a string is given by the formula: \( v = \sqrt{\frac{F}{\mu}} \), where \( F \) is the tension in the rope, and \( \mu = \frac{M}{L} \) is the linear mass density of the rope.

Step 3: Find Klyde's Top Speed (Part a)

Klyde's vertical motion due to the wave can be described by simple harmonic motion. His top speed is given by the maximum transverse speed: \( v_\text{max} = \omega A \), where \( \omega = \frac{2\pi v}{\lambda} \) is the angular frequency of the wave. Substitute \( \omega \) and \( v \) to get \( v_\text{max} = \frac{2\pi A}{\lambda} \sqrt{\frac{F}{\mu}} \).

Step 4: Calculate Adjustment for Tension (Part b)

To double Klyde's top speed, we set \( 2v_\text{max} = \frac{2\pi A}{\lambda} \sqrt{\frac{F_\text{new}}{\mu}} \), where \( F_\text{new} \) is the new tension. Solving for \( F_\text{new} \), we get \( F_\text{new} = 4F \). Thus, the tension should be increased by a factor of 4.

Key Concepts

Transverse Wave

A transverse wave is characterized by oscillations that are perpendicular to the direction of the wave's propagation. Imagine a rope being shaken up and down; each particle of the rope moves vertically while the wave travels horizontally.

• Direction of oscillation: Perpendicular to wave travel.
• Example: Shaking a jump rope.
• Media: Can travel through solid mediums like strings or cables.

In Klyde’s case, the rope moves up and down while the wave travels along its length, carrying Klyde with it. This type of motion does not require Klyde’s mass to influence the wave behavior, allowing the wave to propagate smoothly on the rope.

Sinusoidal Wave

A sinusoidal wave is a smooth, repetitive oscillation that describes wave motion beautifully. It is described mathematically with sine or cosine functions. It is a type of periodic wave often found in various physical phenomena such as light, sound, and mechanical waves on a stretched rope.

• Defined by sine or cosine functions.
• Has a consistent amplitude and wavelength.
• Represents waves in ideal conditions without any damping.

In the exercise, the wave propagating on the rope where Klyde sits is sinusoidal. This means it has a regular pattern, with all the wave properties such as amplitude and wavelength being consistent throughout.

Harmonic Motion

Harmonic motion refers to repetitive movements that follow a set path or cycle. Simple harmonic motion is a subset where the force that returns the object to its equilibrium position is directly proportional to the displacement and acts in the opposite direction.

• Periodic: Movement repeats after equal intervals.
• Described by mathematical sine or cosine functions.
• Example: Pendulums and springs in ideal scenarios.

In this problem, Klyde experiences simple harmonic motion due to the sinusoidal oscillations of the wave. His motion can be calculated using principles of simple harmonic motion, allowing us to determine his top speed moving up and down on the vertical plane.

Angular Frequency

Angular frequency, represented by the symbol \( \omega \), is a measure of how quickly an object moves through its cycle of motion. It's different from the regular frequency by incorporating the factor of \( 2\pi \), which accounts for the wave's circular nature in sinusoidal functions.

• Measured in radians per second.
• Given by the formula \( \omega = \frac{2\pi v}{\lambda} \), where \( v \) is the wave speed and \( \lambda \) is the wavelength.
• Essential for calculating the motion of objects under oscillatory movement.

In Klyde’s journey along the wave, understanding angular frequency is crucial because it helps determine his top speed, shown in the solution as \( v_\text{max} = \omega A \). His movement’s frequency is translated into how fast he oscillates due to the wave’s sinusoidal nature.