/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 An ant with mass \(m\) is standi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An ant with mass \(m\) is standing peacefully on top of a horizontal, stretched rope. The rope has mass per unit length \(\mu\) and is under tension \(F\). Without warning, Cousin Throckmorton starts a sinusoidal transverse wave of wavelength \(\lambda\) propagating along the rope. The motion of the rope is in a vertical plane. What minimum wave amplitude will make the ant become momentarily weightless? Assume that \(m\) is so small that the presence of the ant has no effect on the propagation of the wave.

Short Answer

Expert verified
The minimum wave amplitude (A) that will make the ant weightless can be calculated using the formula \(A = g\lambda/\sqrt{F/\mu}\), where g is the acceleration due to gravity, lambda is the wavelength, F is the tension in the rope, and mu is the mass per unit length of the rope.

Step by step solution

01

Calculate Wave Velocity

Start by calculating the velocity (v) of the wave in terms of the given tension (F) and the mass per unit length (mu). We use the formula for wave speed: \(v = \sqrt{F/\mu}\). This is a standard wave equation derived from Newton's second law.
02

Determine Wave Frequency

The wave frequency (f) is given by the wave velocity (v) divided by the wavelength (lambda). This is derived from the wave equation \(v=f\lambda\). Therefore, f equals to \(v/\lambda\). Use the velocity calculated in Step 1.
03

Develop Relationship between Frequency and Amplitude

The maximum vertical velocity of a point on the wave (and hence the ant) will occur when the motion of the wave is upwards and at its maximum speed. This is when the ant will be weightless. The maximum vertical speed (\(v_y\)) is the product of the frequency (f) and the amplitude (A). Set up the equation: \(v_y=fA\).
04

Equate to Gravity and Solve for Amplitude

To achieve weightlessness, the maximum vertical velocity (\(v_y\)) must equal the acceleration due to gravity (g), hence equate \(fA\) to \(g\). Therefore, \(A = g/f\). Plug the value of frequency (f) from step 2 into this equation to solve for the amplitude (A).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Wave Velocity
Wave velocity, often symbolized as 'v', is a fundamental concept in wave physics, essential for comprehension of how waves travel. It is the speed at which a wave's phase, such as the crest or trough, propagates through space. In the context of our exercise involving an ant and a sinusoidal wave on a rope, the wave velocity was calculated to understand the dynamics of the wave.

Using the formula for wave velocity, \(v = \sqrt{F/\mu}\), we found the relationship between the tension in the rope (F), the mass per unit length (\mu), and how these factors influence the speed of the wave. The tension provides the restoring force that allows the wave to travel, while the mass per unit length acts as inertia, slowing the wave down. The square root in the equation illustrates that the wave speed is not directly proportional to the tension or inversely proportional to the mass per unit length, but it is related to the square of these variables. This means, even if the tension is quadrupled, the wave velocity would only double, demonstrating the delicate balance between the medium's properties and the energy it carries.

This concept is pivotal because it lays the groundwork for discussing other important wave characteristics such as frequency and amplitude.
Unraveling Wave Frequency
In our journey through transverse wave physics, the next stop is wave frequency, denoted by 'f'. Frequency measures how often something happens over a particular time, and in wave terms, it is the number of complete waves, or cycles, passing a point per second. The unit commonly used is hertz (Hz), where 1 Hz equals one cycle per second.

In the given exercise, we deduced the wave frequency from our calculated wave velocity and the wavelength, using the fundamental relationship \(f = v/\lambda\). The wavelength (\lambda) is the distance over which the wave's shape repeats, and it inversely affects the frequency; the longer the wavelength, the lower the frequency, as each cycle takes longer to complete. Understanding wave frequency is crucial, as it directly relates to the energy carried by the wave—the higher the frequency, the higher the energy and vice versa.

In practical terms for the problem at hand, knowing the frequency allowed us to explore how often the ant must oscillate to maintain brief moments of weightlessness, thereby establishing a direct link to the amplitude needed to achieve this effect.
Exploring Wave Amplitude
Wave amplitude is commonly referred to simply as amplitude and is represented by the symbol 'A'. It is perhaps the most visually intuitive aspect of a wave, corresponding to its maximum displacement from the rest position. In our scenario where the ant rides a wave on a rope, the amplitude would be the maximum height the wave reaches, which in turn determines how high the ant gets lifted with each wave cycle.

The amplitude of a wave is a measure of its energy. For mechanical waves, like our sinusoidal wave on a rope, a larger amplitude means more energy is being transferred. This is directly related to the kinetic energy imparted to the ant by the wave—the greater the amplitude, the greater the ability of the wave to lift the ant.

In step 3 of the solution, we're examining when the ant becomes weightless, which is indicative of a balance between the upward force of the wave and the downward force of gravity. By equating the maximum vertical velocity of the wave, a product of frequency and amplitude (\(v_y=fA\)), to the acceleration due to gravity (\(g\)), and subsequently solving for amplitude (\(A = g/f\)), we discover the minimum amplitude needed for the ant to experience momentary weightlessness. This beautifully illustrates the direct consequence amplitude has on the physical experience of the wave's energy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A string or rope will break apart if it is placed under too much tensile stress [see Eq. (11.8)]. Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has density \(7800 \mathrm{~kg} / \mathrm{m}^{3}\) and will break if the tensile stress exceeds \(7.0 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\). You want to make a guitar string from \(4.0 \mathrm{~g}\) of this type of steel. In use, the guitar string must be able to withstand a tension of \(900 \mathrm{~N}\) without breaking. Your job is to determine (a) the maximum length and minimum radius the string can have; (b) the highest possible fundamental frequency of standing waves on this string, if the entire length of the string is free to vibrate.

A horizontal wire is stretched with a tension of \(94.0 \mathrm{~N},\) and the speed of transverse waves for the wire is \(406 \mathrm{~m} / \mathrm{s}\). What must the amplitude of a traveling wave of frequency \(69.0 \mathrm{~Hz}\) be for the average power carried by the wave to be \(0.365 \mathrm{~W} ?\)

A piano wire with mass \(3.00 \mathrm{~g}\) and length \(80.0 \mathrm{~cm}\) is stretched with a tension of \(25.0 \mathrm{~N}\). A wave with frequency \(120.0 \mathrm{~Hz}\) and amplitude \(1.6 \mathrm{~mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is \(0.386 \mathrm{~m}\). The maximum transverse acceleration of a point at the middle of the segment is \(8.40 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\) and the maximum transverse velocity is \(3.80 \mathrm{~m} / \mathrm{s}\). (a) What is the amplitude of this standing wave? (b) What is the wave speed for the transverse traveling waves on this string?

Transverse waves on a string have wave speed \(8.00 \mathrm{~m} / \mathrm{s},\) amplitude \(0.0700 \mathrm{~m},\) and wavelength \(0.320 \mathrm{~m} .\) The waves travel in the \(-x\) -direction, and at \(t=0\) the \(x=0\) end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of a particle at \(x=0.360 \mathrm{~m}\) at time \(t=0.150 \mathrm{~s}\). (d) How much time must elapse from the instant in part (c) until the particle at \(x=0.360 \mathrm{~m}\) next has maximum upward displacement?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.