/*! 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 76 \(\textbf{Longitudinal Waves on ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\textbf{Longitudinal Waves on a Spring.}\) A long spring such as a Slinky\(^{\mathrm{TM}}\) is often used to demonstrate longitudinal waves. (a) Show that if a spring that obeys Hooke’s law has mass \(m\), length \(L\), and force constant \(k'\), the speed of longitudinal waves on the spring is \(v = L\sqrt{ k'/m}\) (see Section 16.2). (b) Evaluate \(v\) for a spring with \(m =\) 0.250 kg, \(L =\) 2.00 m, and \(k' =\) 1.50 N\(/\)m. \(\textbf{ULTRASOUND IMAGING}\). A typical ultrasound transducer used for medical diagnosis produces a beam of ultrasound with a frequency of 1.0 MHz. The beam travels from the transducer through tissue and partially reflects when it encounters different structures in the tissue. The same transducer that produces the ultrasound also detects the reflections. The transducer emits a short pulse of ultrasound and waits to receive the reflected echoes before emitting the next pulse. By measuring the time between the initial pulse and the arrival of the reflected signal, we can use the speed of ultrasound in tissue, 1540 m/s, to determine the distance from the transducer to the structure that produced the reflection. As the ultrasound beam passes through tissue, the beam is attenuated through absorption. Thus deeper structures return weaker echoes. A typical attenuation in tissue is \(-\)100 dB/m \(\cdot\) MHz; in bone it is \(-\)500 dB/m \(\cdot\) MHz. In determining attenuation, we take the reference intensity to be the intensity produced by the transducer.

Short Answer

Expert verified
The speed of longitudinal waves on the spring is approximately 4.90 m/s.

Step by step solution

01

Understand Hooke's law for spring

Hooke's law states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, this is expressed as \( F = k' \Delta L \), where \( F \) is the force applied, \( k' \) is the spring constant, and \( \Delta L \) is the change in the spring's length.
02

Derive wave speed formula for spring

For a spring, the speed of longitudinal waves depends on the tension in the spring and the mass per unit length of the spring. The wave speed \( v \) is given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension and \( \mu = \frac{m}{L} \) is the linear mass density. Since for a spring under a stretch, tension \( T \) is given by \( k'L \), we have: \[ v = \sqrt{\frac{k'L}{m/L}} = L\sqrt{\frac{k'}{m}}. \] This is the derived formula for the wave speed.
03

Substitute given values to calculate wave speed

Using the derived formula \( v = L\sqrt{\frac{k'}{m}} \), substitute \( L = 2.00 \text{ m} \), \( m = 0.250 \text{ kg} \), and \( k' = 1.50 \text{ N/m} \): \[ v = 2.00\sqrt{\frac{1.50}{0.250}}. \] Calculate the expression to find \( v \).
04

Calculate numerical value

Perform the calculations: \( \sqrt{\frac{1.50}{0.250}} = \sqrt{6} = 2.45 \) (approximately). Hence, \( v = 2.00 \times 2.45 \approx 4.90 \text{ m/s} \).

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.

Hooke's Law
Hooke's Law is a fundamental principle that describes how springs behave when they are stretched or compressed. According to this law, the force required to change the length of a spring is directly proportional to the change in length. This can be described using the formula:
  • \( F = k' \Delta L \)
where:
  • \( F \) is the force applied,
  • \( k' \) is the spring constant (specific to each spring), and
  • \( \Delta L \) is the change in length from the spring's natural state.

This law is crucial in understanding many mechanical systems, from simple toys to complex engineering designs. By knowing the spring constant \( k' \), we can predict how much force is needed to compress or extend a spring to a given length. This concept is not only used for springs; any object that deforms in a linear fashion may be analyzed using Hooke’s Law.
Using this law, you can also see why springs can "store" energy – as you apply force and change the spring's length, potential energy is stored within the spring, ready to be released when the spring returns to its original shape.
Wave Speed
Wave speed in the context of longitudinal waves on a spring refers to how fast the disturbance travels through the spring. This is determined by both the tension in the spring and the linear mass density. The general formula for calculating wave speed \( v \) is:
  • \( v = \sqrt{\frac{T}{\mu}} \)
where:
  • \( T \) is the tension in the spring, often calculated using Hooke’s Law as \( T = k'L \), and
  • \( \mu \) is the mass per unit length of the spring, given by \( \mu = \frac{m}{L} \).

By substitution, the specific formula for a spring becomes \( v = L\sqrt{\frac{k'}{m}} \). This shows that wave speed increases with the spring constant and decreases with the mass of the spring.
In practical scenarios, knowing the wave speed allows you to understand how quickly signals or energy can be transmitted through the spring system. This concept is widely applicable in various fields, including mechanical waves in materials, sound waves, and even electrical signals that mimic wave-like properties.
Ultrasound Imaging
Ultrasound imaging is a fascinating application of wave principles, specifically using sound waves with frequencies higher than the normal human hearing range. Medical ultrasound usually operates at frequencies around 1 MHz. These high-frequency waves are emitted and detected by a device called a transducer.
During an ultrasound scan, the transducer sends out short pulses of ultrasound waves into the body. These waves travel through tissues and are partially reflected back when they hit a boundary between different types of tissue. By measuring the time taken for the waves to return, the ultrasound machine calculates the distance to the reflective surfaces (anatomy) using the known speed of sound in tissue, typically 1540 m/s.
  • Attenuation: As ultrasound waves travel through the body, they lose intensity—this is called attenuation. Different tissues attenuate ultrasound waves differently, with typical values being -100 dB/m MHz for tissue and -500 dB/m MHz for bone.
  • Echoes: The reflected waves, or echoes, are received by the transducer and used to create images of the organs and tissues inside the body.
Ultrasound imaging is non-invasive and safe, making it an invaluable tool for medical diagnostics, especially in monitoring fetal development and evaluating soft tissues.

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

You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 4110 N. You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second \(overtone\) with very large amplitude. How long should the pipe be?

Horseshoe bats (genus \(Rhinolophus\)) emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A \(Rhinolophus\) flying at speed \(v_{bat}\) emits sound of frequency \(f_{bat}\); the sound it hears reflected from an insect flying toward it has a higher frequency \(f_{refl}\). (a) Show that the speed of the insect is $$vinsect = v\Bigg[\frac{f_{refl}(v - v_{bat}) - f_{bat}(v + v_{bat})}{f_{refl}(v - v_{bat}) + f_{bat}(v + v_{bat})}\Bigg] $$ where \(v\) is the speed of sound. (b) If \(f_{bat} =\) 80.7 kHz, \(f_{refl} =\) 83.5 kHz, and \(v_{bat} =\) 3.9 m/s, calculate the speed of the insect.

A soprano and a bass are singing a duet. While the soprano sings an A-sharp at 932 Hz, the bass sings an A-sharp but three octaves lower. In this concert hall, the density of air is 1.20 kg/m\(^3\) and its bulk modulus is 1.42 \(\times\) 10\(^5\) Pa. In order for their notes to have the same sound intensity level, what must be (a) the ratio of the pressure 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 m and in nm) does the soprano produce to sing her A-sharp at 72.0 dB?

While sitting in your car by the side of a country road, you are approached by your friend, who happens to be in an identical car. You blow your car's horn, which has a frequency of 260 Hz. Your friend blows his car's horn, which is identical to yours, and you hear a beat frequency of 6.0 Hz. How fast is your friend approaching you?

The sound source of a ship's sonar system operates at a frequency of 18.0 kHz. The speed of sound in water (assumed to be at a uniform 20\(^\circ\)C) is 1482 m/s. (a) What is the wavelength of the waves emitted by the source? (b) What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling directly toward the ship at 4.95 m/s ? The ship is at rest in the water.

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.