/*! 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 53 A non-uniform wire of length \(L... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A non-uniform wire of length \(L\) and mass \(M\) has a variable density given by \(\mu=k x\) where \(x\) is the distance from one end of the wire and \(k\) is a constant. Find the time required for a pulse gencrated at one end of the wire to travel to the other end when tension in the wirc is \(T\).

Short Answer

Expert verified
The time for the pulse to travel is \( \frac{2}{3} \sqrt{\frac{2ML}{T}} \).

Step by step solution

01

Expression for Mass Element

The density of the wire is given by \( \mu = kx \). The mass of an infinitesimally small element of the wire, \( dm \), at a distance \( x \) from one end, is expressed as \( dm = \mu \, dx = kx \, dx \).
02

Find Total Mass of the Wire

To find the total mass, we integrate the expression for \( dm \) over the entire length of the wire: \[ M = \int_0^L kx \, dx = \left[ \frac{kx^2}{2} \right]_0^L = \frac{kL^2}{2}.\]
03

Determine the Constant k

We are given that the total mass is \( M \). Therefore, we can set the expression from the previous step equal to \( M \) to solve for \( k \):\[ M = \frac{kL^2}{2} \implies k = \frac{2M}{L^2}. \]
04

Speed of Pulse on the Wire

The speed \( v \) of a pulse on a wire is given by \( v = \sqrt{\frac{T}{\mu}} \). Substituting for \( \mu = kx \), it becomes \( v = \sqrt{\frac{T}{kx}} \).
05

Expression for Time

The time \( dt \) taken for a pulse to travel an infinitesimal distance \( dx \) is given by \( dt = \frac{dx}{v} = \frac{dx}{\sqrt{\frac{T}{kx}}} = \sqrt{\frac{kx}{T}} \, dx \).
06

Integrate to Find Total Time

To find the total time \( t \) required for the pulse to travel from \( x = 0 \) to \( x = L \), we integrate the expression for \( dt \):\[t = \int_0^L \sqrt{\frac{kx}{T}} \, dx = \sqrt{\frac{k}{T}} \int_0^L \sqrt{x} \, dx. \]
07

Compute the Integral

The integral \( \int_0^L \sqrt{x} \, dx \) can be solved as follows: \[\int_0^L \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^L = \frac{2}{3} L^{3/2}. \]
08

Substitute k and Simplify

Substitute \( k = \frac{2M}{L^2} \) from Step 3 into the expression for \( t \):\[t = \sqrt{\frac{\frac{2M}{L^2}}{T}} \times \frac{2}{3} L^{3/2} = \frac{2}{3} \sqrt{\frac{2ML}{T}}. \]

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.

Variable Density Wire
A variable density wire is a fascinating concept where the density is not consistent throughout the entire length. In the given problem, the density of the wire changes as you move along its length according to the function \( \mu = kx \), where \( x \) is the distance from one end of the wire and \( k \) is a constant. This means as you move further along the wire, its density increases linearly. Understanding how density varies can aid in predicting how other physical properties, like mass distribution and wave propagation, are affected by non-uniformity.

This concept is particularly useful in physics where practical applications involve materials with varying compositions. For wires or strings, knowing the exact distribution of density is key when analyzing the motion of waves, as well as calculating other important parameters such as tension and speed.
Integration in Physics
Integration is a powerful mathematical tool used extensively in physics to deal with continuous variables. When a physical quantity, such as mass or charge, is distributed over a region, integration helps find the total value of that quantity. In the context of this exercise, integration allows us to calculate the total mass of the wire, where the density varies along its length as \( \mu = kx \).

The integration process involves summing up infinitesimal elements over the desired range. For the wire, we calculated the mass of each small element \( dm = kx \, dx \) and then integrated this expression from 0 to \( L \) to find the total mass \( M \). This method of integration provides essential insight into non-uniform systems and is arguably one of the fundamental techniques in physics for solving problems involving variable distributions.
Speed of Mechanical Waves
The speed of mechanical waves in a medium, like a wire or string, depends significantly on two factors: the tension in the medium \( T \) and its linear density \( \mu \). The formula \( v = \sqrt{\frac{T}{\mu}} \) shows that wave speed increases with higher tension and decreases with higher density.

For our non-uniform wire, where the density \( \mu \) varies as \( \mu = kx \), the wave speed becomes a function of position \( x \): \( v = \sqrt{\frac{T}{kx}} \). This relationship implies that as you move away from the denser end of the wire, the speed of the wave varies. Understanding how different factors affect wave speed is crucial, especially in engineering and acoustics, where precise control of wave propagation is needed.
Tension in Wire
Tension in a wire is a force that is exerted along the length of the wire, keeping it taut. It plays a vital role in determining how waves propagate through the wire. In this exercise, tension \( T \) is a constant force applied to the wire, affecting both the speed of the wave and the time it takes for a pulse to travel from one end of the wire to the other.

Tension can be influenced by factors like the weight of the wire or external forces ensuring it remains stretched. Higher tension generally means that a wave will travel faster through the wire, as seen in our speed equation \( v = \sqrt{\frac{T}{\mu}} \). Recognizing how tension affects wave behavior is foundational in physics, often applicable in designing musical instruments, calculating structural loads, and understanding seismic activity.

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

In the state of rotation of the system as a whole there are two forces which act on the sphere: (i) elastic and (ii) centripetal. The equation of motion of the sphere is \(\frac{m d^{2} x}{d t^{2}}=-k x+m \omega^{2} x ;\) (here \(x\) sunds for displacement) or \(\frac{d^{2} x}{d t^{2}}=-\left(\frac{k}{m}-\omega^{2}\right) x\) i.e., the sphere will execute S.H.M. with $$ \begin{aligned} \omega_{n} &=\sqrt{\frac{k}{m}-\omega^{2}} \\ \therefore \quad T &=\frac{2 \pi}{\omega_{o}}=\frac{2 \pi}{\sqrt{k / m-\omega^{2}}} \\ &=\frac{2 \times 3.14}{\sqrt{\frac{20}{0.2}-(4.4 \times 4.4)}} \approx 0.7 \mathrm{~s} \end{aligned} $$ The oscillation will stop when \(T=\infty\), This is posible when $$ \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{20}{0.2}}=10 \mathrm{rad} / \mathrm{s} $$

The frequency of an organ pipe at \(15^{\circ} \mathrm{C}\) is \(156 \mathrm{~Hz}\). At what temperature, will the frequency be \(160 \mathrm{~Hz}\) ? neglect expansion of tube.

There is a set of 4 tuning forks, one with lowest frequency vibrating at \(552 \mathrm{~Hz}\). By using any two forks at a time, the beat frequencies heard are \(1,2,3,5,7,8 .\) The possible frequencies of other three forks are (a) 553,554 and \(560 \mathrm{~Hz}\) (b) 553,555 and \(560 \mathrm{~Hz}\) (c) 553,556 and \(558 \mathrm{~Hz}\) (d) 551,554 and \(560 \mathrm{~Hz}\)

A string of length \(L\) is stretched along the \(x\)-axis and is rigidly clamped at its two ends. It undergoes transverse vibrations. If \(n\) is an integer, which of the followings relations may represent the shape of the string at any time \(t ?\) (a) \(y-\Lambda \sin \left(\frac{n \pi x}{L}\right) \cos \omega\) (b) \(y-\Lambda \sin \left(\frac{n \pi x}{L}\right) \sin \alpha\) (c) \(y-A \cos \left(\frac{n \pi x}{L}\right) \cos \omega\) (d) \(y-A \cos \left(\frac{n \pi x}{L}\right) \sin \omega\)

Statement-1: If amplitude of SIIM is doubled, the time period remains same. Statement-2: \(\Lambda\) amplitude and periodicity are two independent characteristics of SIIM.

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.