/*! 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 35 A wave pulse starts propagating ... [FREE SOLUTION] | 91Ó°ÊÓ

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A wave pulse starts propagating in the positive \(x\)-direction along a non- uniform wire of length \(10 \mathrm{~m}\) with a mass per unit length give by \(m=m_{\mathrm{g}}+\alpha x\) and under a tension of \(100 \mathrm{~N}\). Find the time taken by the pulse to travel from the lighter end \((x=0)\) to the heavier end. $$ \left(m_{n}=10^{-2} \mathrm{~kg} / \mathrm{m} \text { and } \alpha=9 \times 10^{-3} \mathrm{~kg} / \mathrm{m}^{2}\right) $$

Short Answer

Expert verified
The time taken by the wave pulse is approximately \(1.133 \times 10^{-4}\) seconds.

Step by step solution

01

Understand the Problem

We need to find the time taken by a wave pulse to travel along a non-uniform wire. The wire has a mass per unit length that changes with position along the wire as given by the formula \(m = m_g + \alpha x\). The tension in the wire is constant at 100 N.
02

Determine the Wave Speed Expression

The speed \(v\) of a wave on a string is given by the formula \(v = \sqrt{\frac{T}{\mu}}\), where \(\mu\) is the linear mass density and \(T\) is the tension. Given \(T = 100\, \text{N}\) and \(\mu = m_g + \alpha x\). Thus, \(v(x) = \sqrt{\frac{100}{m_g + \alpha x}}\).
03

Set Up the Integral for Travel Time

The time \(t\) taken for a wave pulse to travel from \(x=0\) to \(x=10\, \text{m}\) is given by the integral \(t = \int_{0}^{10} \frac{dx}{v(x)}\). Substituting \(v(x) = \sqrt{\frac{100}{m_g + \alpha x}}\), the integral becomes \(t = \int_{0}^{10} \frac{\sqrt{m_g + \alpha x}}{10} \, dx\).
04

Simplify the Integral Expression

First, express the integrand as \(\frac{\sqrt{m_g + \alpha x}}{10}\). This simplifies to \(\frac{1}{10} \cdot (m_g + \alpha x)^{1/2}\).
05

Evaluate the Integral

The integral \(t = \frac{1}{10} \int_{0}^{10} (m_g + \alpha x)^{1/2} \, dx\) is a standard form integral. Let \(u = m_g + \alpha x\). Then, \(du = \alpha dx\) or \(dx = \frac{du}{\alpha}\). The integral becomes:\[ t = \frac{1}{10\alpha} \int_{m_g}^{m_g + 10\alpha} u^{1/2} \, du \].
06

Apply the Power Rule for Integration

Evaluate the integral \(\int u^{1/2} \, du\) which is \(\frac{2}{3} u^{3/2} + C\). Therefore, the time \(t\) is:\[ t = \frac{1}{30\alpha} [ (m_g + 10\alpha)^{3/2} - m_g^{3/2} ] \].
07

Substitute the Given Values

Substitute \(m_g = 10^{-2}\, \text{kg/m}\) and \(\alpha = 9 \times 10^{-3}\, \text{kg/m}^2\) into the expression:\[ t = \frac{1}{270} [ (10^{-2} + 9 \times 10^{-2})^{3/2} - (10^{-2})^{3/2} ] \].
08

Calculate the Results

Calculate the expression:1. \((10^{-2} + 9 \times 10^{-2}) = 10^{-1}\), \((10^{-1})^{3/2} = 10^{-3/2} = 0.0316\).2. \((10^{-2})^{3/2} = (10^{-3}) = 0.001\).3. Substitute the values: \(t = \frac{1}{270} (0.0316 - 0.001) = \frac{1}{270} \cdot 0.0306\).4. Final calculation: \(t \approx 1.133 \times 10^{-4}\, \text{seconds}\).

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Key Concepts

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

Non-uniform Wire
A non-uniform wire is a wire that does not have a constant mass per unit length. This means the wire's density varies along its length. In our exercise, the mass changes as you move from one end of the wire to the other. It is defined by the formula \( m = m_g + \alpha x \), where \( m_g \) is the mass per unit length at the starting point and \( \alpha \) is a factor indicating how much the mass per unit length increases linearly with distance \( x \).
This characteristic is crucial when considering wave propagation because the changing mass impacts how quickly the wave can travel through different sections of the wire.
  • Non-uniform means not consistent or not the same across its length.
  • The variable mass affects calculations in determining wave speed.
Understanding this property helps us model more complex real-world situations where uniformity is rare.
Mass Per Unit Length
Mass per unit length, often denoted as \( \mu \), is a measure that tells us how much mass is contained in a specific length of the wire. In problems involving wave propagation, determining the mass per unit length is essential because it directly affects the wave speed and behavior.
In our problem scenario, mass per unit length is not a constant, but instead, it varies according to the equation \( \mu = m_g + \alpha x \). This is significant because both \( m_g \) and \( \alpha x \) contribute to the overall mass at any point \( x \) along the wire.
  • Helps in understanding how mass is distributed along a wire or a string.
  • A key component in determining wave speed which influences how waves travel.
  • Integral calculus is used to accommodate this variability in our calculations.
Recognizing this variation is critical for accurate calculations of wave speed on non-uniform wires.
Wave Speed
Wave speed is a fundamental concept when discussing wave propagation. It is defined as the speed at which a wave travels through a medium. The speed of a wave on a string is given by the formula:
  • \( v = \sqrt{\frac{T}{\mu}} \)
Where \( v \) is the wave speed, \( T \) is the tension in the string, and \( \mu \) is the mass per unit length of the string.
As the mass per unit length varies along the wire, as it does in our exercise, the wave speed also changes accordingly. This variable speed needs careful calculation since different sections of the wire carry different masses.
Understanding wave speed in variable conditions is crucial for engineering and physics, as waves in real-world materials rarely encounter uniform conditions.
Integral Calculus
Integral calculus is powerful for dealing with quantities that vary across a continuum. In our exercise, we used integral calculus to find the time it takes for a wave to travel along the non-uniform wire. The equation \( t = \int_{0}^{10} \frac{dx}{v(x)} \) represents the formula for calculating the travel time along the wire using the changing wave speed \( v(x) \).
This integral accounts for the varying mass per unit length, where the wave speed changes with \( x \). Calculating such integrals often involves substitutions, which simplify the math. For instance, by setting \( u = m_g + \alpha x \) we make the integral more manageable, allowing us to apply the power rule for evaluation.
  • Used to sum continuous changes along a path.
  • Helps in determining variable-dependent values over a defined range.
  • Integral calculus provides tools for precise analysis of non-uniform systems.
Mastering integral calculus is necessary for tackling complex physics problems involving dynamic systems.

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Most popular questions from this chapter

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Statement- \(1: \Lambda\) particle moves on \(x\)-axis according to equation \(x=2+6 \sin (\pi t)\). Motion is nol SIIM. Statement-2 : The acceleration is not proportional to displacement from origin.

A disc of radius \(20 \mathrm{~m}\) is rotating uniformly with angular frequency \(\omega=10 \mathrm{rad} / \mathrm{s}\), A source is fixod to rim of dise. The ratio of maximum and minimum froquency heard by observer far away from disc in plane of disc is (take speed of sound \(330 \mathrm{~m} / \mathrm{s}\) ) (a) \(33 / 13\) (b) \(33 / 53\) (c) \(13 / 53\) (d) \(53 / 13\)

Statement-1: When a source of sound is approaching an observer, the frequency perceived by the observer is greater than the frequency of the sound source. Statement-2: As the sound-source approaches the listener, the wavelength of the sound decreases while its speed relative to the medium remains unchanged.

(a) Let \(L \rightarrow\) stationary, \(\nu=580 \mathrm{II} \mathrm{z}\) \(s \rightarrow\) moving with \(40 \mathrm{~km} / \mathrm{hr}\) and aproaching \(w \rightarrow 40 \mathrm{~km} / \mathrm{hr}\) and supporting \(v=1200 \mathrm{~km} / \mathrm{hr}\) So \(v^{\prime}=\left[\frac{v+\omega}{(v+\omega)-v_{s}}\right] v\) \(\Rightarrow \quad v^{\prime}=\frac{1240}{1200} \times 580=\frac{1798}{3}=599.33 \mathrm{IIz}\) (b) At a distance \(1 \mathrm{~km}\) before the train whistles and let the driver heard the echo at time \(l=t^{\prime}\). then at this time, total distance travelled by wave and train \(=2 \mathrm{~km}\) Now time after which first wave reach to the hill in time interval \(t_{1}=\frac{1}{1200} \mathrm{hr}\) after reflecting at \(t=t^{\prime}\), echo will heard, then distance travellcd by it is \(x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200\) and distance travelled by the train is \(x_{2}=\left(t^{\prime}-\frac{1}{1200}\right) 40\) As \(x_{1}+x_{2}=1 \mathrm{~km}\) \(\therefore\left(t^{\prime}-\frac{1}{1200}\right) \times 12040+\left(t^{\prime}-\frac{1}{1200}\right) 40\) \(1240 t^{\prime}=\frac{1}{30} \quad \therefore \quad t^{\prime}=\frac{61}{30 \times 1240}\) and required distance, $$ x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200 $$$=\left(\frac{61}{30 \times 1240}-\frac{1}{1200}\right) \times 1200=\frac{61 \times 1200}{30 \times 1240}-1\( \)=\frac{30}{31} \times 1000 \mathrm{~m}=967.74 \mathrm{~m}\( Frequency of sound heard by the driver, \)v^{\prime \prime}=\left(\frac{(v-\omega)+v_{2}}{v-\omega}\right) v\( \)=\left(\frac{1200-40+40}{1200-40}\right) \times 599.33\( \)\therefore \quad v^{\prime \prime}=\frac{1200}{1160} \times 599.33=620 \mathrm{~Hz}$

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