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A long spring such as a Slinkyrs 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 \(k^{\prime}\), the speed of longitudinal waves on the spring is \(v=L \sqrt{k^{\prime} / m}\) (see Section 16.2 ). (b) Evaluate \(v\) for a spring with \(m=0.250 \mathrm{~kg}, L=2.00 \mathrm{~m},\) and \(k^{\prime}=1.50 \mathrm{~N} / \mathrm{m}\)

Short Answer

Expert verified
The speed of longitudinal waves on a spring that obeys Hooke's law is given by \(v=L \sqrt{k^{\prime} / m}\). For a spring with specified parameters \(m=0.250 \mathrm{~kg}, L=2.00 \mathrm{~m},\) and \(k^{\prime}=\1.50 \mathrm{~N} / \mathrm{m}\), the speed of longitudinal waves on the spring is \(\sqrt{6}\) m/s.

Step by step solution

01

Derive the wave speed formula

Firstly, recall Hooke's law for springs, \(F = -k'x\), where \(F\) is the force, \(k'\) is the spring constant, and \(x\) is the displacement of the spring. This shows that the force to stretch or compress the spring is proportional to the displacement from its equilibrium point. Also, recall the formula for wave speed \(v\) on a string stretched with tension \(T\) and total mass \(M\), \(v = \sqrt{T/M}\). For a longitudinal wave on the spring, the tension (force) in the spring is \(k'x\), and so substituting this into the wave speed formula gives \(v = \sqrt{k'x/M}\). Furthermore, if the total mass of the spring is \(m\), then \(M = m/L\) the mass per unit length. Substituting this into the wave speed formula gives \(v = \sqrt{k'L/m}\).
02

Evaluate the wave speed

Given \(m=0.250 \mathrm{~kg}, L=2.00 \mathrm{~m},\) and \(k'=1.50 \mathrm{~N} / \mathrm{m}\), substitute these values into \(v = \sqrt{k'L/m}\) to evaluate the speed of longitudinal waves on the spring. \(v = \sqrt{(1.50 \mathrm{~N / m})(2.00 \mathrm{~m}) / 0.250 \mathrm{~kg}}\). This calculation yields \(v = \sqrt{6}\) m/s. Therefore, the speed of longitudinal waves on the spring is \(\sqrt{6}\) m/s.

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Key Concepts

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

Hooke's Law
Understanding Hooke's law is essential when studying the behavior of springs and their interactions with forces. In the realm of physics, Hooke's law is a fundamental principle that defines the relationship between the force applied to a spring and the amount of stretch or compression it experiences. Mathematically, this law is expressed as \( F = -k'x \), where \( F \) represents the force applied, \( k' \) is the spring constant indicating the stiffness of the spring, and \( x \) is the change in length from the spring’s original position.
In simpler terms, the negative sign shows that the force exerted by the spring is in the opposite direction to the displacement. This law tells us that the more you stretch or compress a spring, the more force it will exert to return to its original length. It’s significant to note that Hooke's law is valid only within the elastic limit—when the material returns to its original shape after the force is removed.
Wave Speed Formula
When dealing with wave motion, the wave speed formula is a fundamental concept that allows one to calculate how fast the wave is traveling. For a string or a spring, the wave speed \( v \) is given by the formula \( v = \sqrt{T/M} \), where \( T \) is the tension in the string or spring, and \( M \) is the mass per unit length.
It's crucial to understand that in the context of a stretched spring, the tension is the force responsible for restoring the spring to its equilibrium position, which according to Hooke's law is \( k'x \) for small displacements \( x \). However, for a continuous medium like a long spring used to demonstrate longitudinal waves, \( T \) is equivalent to the spring constant \( k' \) as long as the spring is not significantly compressed or stretched. Therefore, in the adapted wave speed formula specifically for longitudinal waves on a spring, \( v \) changes to \( v = \sqrt{k'/M} \) where \( M \) is the mass per unit length of the spring.
This demonstrates the direct correlation between the stiffness of the spring and the wave speed: stiffer springs (higher \( k' \) values) will transmit waves faster. Conversely, heavier springs (higher \( M \) values) will transmit waves more slowly.
Longitudinal Wave Speed Calculation
The calculation of longitudinal wave speed on a spring hinges on the interplay between Hooke's law and the wave speed formula, as seen in the problem provided. For a longitudinal wave moving along a spring which obeys Hooke’s law, the speed \( v \) is calculated by \( v = L \sqrt{k'/m} \), where \( L \) is the length of the spring, \( k' \) is the spring constant, and \( m \) is the total mass of the spring.
To calculate the wave speed, you must first understand that the mass per unit length (\( M \) in the wave speed formula) is the total mass \( m \) divided by the total length \( L \) of the spring. By substituting \( M \) with \( m/L \) into the wave speed formula, we arrive at the modified expression: \( v = \sqrt{k'L/m} \). This calculation allows us to predict how quickly a wavetrain will travel along a spring, assuming it follows Hooke's law.
In practical applications—such as engineering, acoustics, and materials science—knowing how to calculate the speed of longitudinal waves is critically important since it affects how structures respond to stress and vibration. Understanding these concepts helps in designing systems and materials that are safe, efficient, and fit-for-purpose.

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Most popular questions from this chapter

A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you can hear them if you are inside the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda_{n}=2 L / n\) and the frequencies are given by \(f_{n}=m v / 2 L=n f_{1},\) where \(n=1,2,3, \ldots\) (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower \(2.50 \mathrm{~m}\) tall. Are these frequencies audible?

You blow across the open mouth of an empty test tube and produce the fundamental standing wave in the \(14.0-\mathrm{cm}\) -long air column in the test tube, which acts as a stopped pipe. (a) What is the frequency of this standing wave? (b) What is the frequency of the fundamental standing wave in the air column if the test tube is half filled with water?

Many professional singers have a range of \(2 \frac{1}{2}\) octaves or even greater. Suppose a soprano's range extends from A below middle C (frequency \(220 \mathrm{~Hz}\) ) up to E-flat above high \(\mathrm{C}\) (frequency \(1244 \mathrm{~Hz}\) ). Although the vocal tract is complicated, we can model it as a resonating air column, like an organ pipe, that is open at the top and closed at the bottom. The column extends from the mouth down to the diaphragm in the chest cavity. Assume that the lowest note is the fundamental. How long is this column of air if \(v=354 \mathrm{~m} / \mathrm{s} ?\) Does your result seem reasonable, on the basis of observations of your body?

A \(2.00 \mathrm{MHz}\) sound wave travels through a pregnant woman's abdomen and is reflected from the fetal heart wall of her unborn baby. The heart wall is moving toward the sound receiver as the heart beats. The reflected sound is then mixed with the transmitted sound, and 72 beats per second are detected. The speed of sound in body tissue is \(1500 \mathrm{~m} / \mathrm{s}\). Calculate the speed of the fetal heart wall at the instant this measurement is made.

Standing sound waves are produced in a pipe that is \(1.20 \mathrm{~m}\) long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the displacement nodes and the pressure nodes if (a) the pipe is open at both ends and (b) the pipe is closed at the left end and open at the right end.

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