/*! 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 6 A small bead of mass \(4.00 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 6

A small bead of mass \(4.00 \mathrm{~g}\) is attached to a horizontal string. Transverse waves of amplitude \(A=0.800 \mathrm{~cm}\) and frequency \(f=20.0 \mathrm{~Hz}\) are set up on the string. Assume the mass of the bead is small enough that the bead doesn't alter the wave motion. During the wave motion, what is the maximum vertical force that the string exerts on the bead?

Short Answer

Expert verified
The maximum vertical force that the string exerts on the bead is \(1.6蟺虏 N\).

Step by step solution

01

Calculate Angular Frequency

The angular frequency 蠅 can be calculated by the formula \(蠅 = 2蟺f\). Here \(f = 20.0 Hz\). Substituting these values, we get \(蠅 = 2蟺 \cdot 20.0 = 40蟺 rad/s\).
02

Calculate Maximum Acceleration

The maximum acceleration \(a_{max}\) can be calculated by the formula \(a_{max} = 蠅^2 \cdot A\). The amplitude \(A = 0.800 cm = 0.008 m\). Substituting these values, we get \(a_{max} = (40蟺)虏 \cdot 0.008 = 400蟺虏 m/s虏\).
03

Calculate Maximum Force

The maximum force \(F_{max}\) can be calculated by the formula \(F_{max} = m \cdot a_{max}\). The mass \(m = 4.00g = 0.004 kg\). Substituting these values we get \(F_{max} = 0.004 \cdot 400蟺虏 = 1.6蟺虏 N\).

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 91影视!

Key Concepts

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

Angular Frequency
Angular frequency is a crucial concept when analyzing oscillatory systems, particularly in physics. It specifies how fast the wave oscillates in radians per second. This is different from regular frequency, which is measured in cycles per second or Hertz (Hz). To calculate angular frequency, you use the formula
  • \( \omega = 2\pi f \)
where \( f \) is the frequency.
For our context, if you have a wave with a frequency of 20 Hz, you multiply it by \( 2\pi \) to convert it into radians per second.
This becomes \( \omega = 40\pi \text{ rad/s} \), allowing you to determine other aspects of the wave, like acceleration and forces acting on objects involved in the motion.
Transverse Waves
Transverse waves are a type of wave where the movement of the medium is perpendicular to the direction of the wave's motion. That means if the wave moves horizontally, the medium (like a string or water) moves up and down.
This characteristic of transverse waves is crucial in many areas, including light and electromagnetic waves, which are transverse in nature.
  • Key features to remember include:
  • The medium oscillates at right angles to the direction of energy transfer.
  • Amplitude is the maximum displacement from the rest position.
  • Wavelength is the distance between consecutive crests or troughs.
These features are essential when calculating the forces and motion in systems like the horizontal string supporting a mass.
Maximum Acceleration
Maximum acceleration in wave motion indicates how quickly the speed of a point on the wave can change. It is valuable for understanding both the dynamic forces involved and how they might affect objects in or on the wave.
To find this, we use the formula:
  • \( a_{\text{max}} = \omega^2 \times A \)
where \( \omega \) is the angular frequency and \( A \) is the amplitude.
Using the earlier calculated \( \omega = 40\pi \) rad/s and the amplitude \( A = 0.008 \) m:
  • The maximum acceleration becomes \( a_{\text{max}} = (40\pi)^2 \times 0.008 = 400\pi^2 \) m/s虏.
This value helps determine the maximum vertical force exerted on an object at the wave's peak acceleration.
Vertical Force Calculation
When considering waves, it's essential to calculate the force acting on objects involved. For vertical force calculation, especially with an object attached to a wave like our bead, you consider the mass and the system's maximum acceleration.
The force is determined using Newton鈥檚 Second Law, expressed as:
  • \( F_{\text{max}} = m \times a_{\text{max}} \)
where \( m \) is the mass and \( a_{\text{max}} \) is maximum acceleration.
In this exercise, the given mass is 4 grams or 0.004 kg, and the maximum acceleration was found as \( 400\pi^2 \) m/s虏:
  • Thus, \( F_{\text{max}} = 0.004 \times 400\pi^2 = 1.6\pi^2 \) N.
This force reflects how strongly the wave could potentially impact the bead at its peak.

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

Standing waves on a wire are described by Eq. (15.28), with \(A_{\mathrm{SW}}=2.50 \mathrm{~mm}, \omega=942 \mathrm{rad} / \mathrm{s},\) and \(k=0.750 \pi \mathrm{rad} / \mathrm{m} .\) The left end of the wire is at \(x=0 .\) At what distances from the left end are (a) the nodes of the standing wave and (b) the antinodes of the standing wave?

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(M g\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu .\) You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M ;\) then you measure the node-to-node distance \(d\) for each standing-wave pattern and obtain the following data: $$ \begin{array}{l|lllll} \text { String } & \text { A } & \text { A } & \text { B } & \text { B } & \text { C } \\ \hline \mu(\mathrm{g} / \mathrm{cm}) & 0.0260 & 0.0260 & 0.0374 & 0.0374 & 0.0482 \\ M(\mathrm{~g}) & 559 & 249 & 365 & 207 & 262 \\ d(\mathrm{~cm}) & 48.1 & 31.9 & 32.0 & 24.2 & 23.8 \end{array} $$ (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^{2}(\) in \(\mathrm{kg} \cdot \mathrm{m})\) versus \(M(\) in \(\mathrm{kg}) .\) Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straightline fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). (d) For string A \((\mu=0.0260 \mathrm{~g} / \mathrm{cm}),\) what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of \(24.0 \mathrm{~cm}\) ? Use the value of \(f\) that you calculated in part (c).

A piano tuner stretches a steel piano wire with a tension of \(800 \mathrm{~N}\). The steel wire is \(0.400 \mathrm{~m}\) long and has a mass of \(3.00 \mathrm{~g}\). (a) What is the frequency of its fundamental mode of vibration? (b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to \(10,000 \mathrm{~Hz} ?\)

A musician tunes the C-string of her instrument to a fundamental frequency of \(65.4 \mathrm{~Hz}\). The vibrating portion of the string is \(0.600 \mathrm{~m}\) long and has a mass of \(14.4 \mathrm{~g}\). (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from \(65.4 \mathrm{~Hz}\) to \(73.4 \mathrm{~Hz}\), corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D}\) ?

A \(1.50-\mathrm{m}\) -long rope is stretched between two supports with a tension that makes the speed of transverse waves \(62.0 \mathrm{~m} / \mathrm{s}\). What are the wavelength and frequency of (a) the fundamental; (b) the second overtone; (c) the fourth harmonic?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Turning Points in Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Force

Read Explanation

Energy Physics

Read Explanation

Electricity and Magnetism

Read Explanation
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.