/*! 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 21 A \(100 \mathrm{~g}\) wire is he... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(100 \mathrm{~g}\) wire is held under a tension of \(250 \mathrm{~N}\) with one end at \(x=0\) and the other at \(x=10.0 \mathrm{~m}\). At time \(t=0\), pulse 1 is sent along the wire from the end at \(x=10.0 \mathrm{~m}\). At time \(t=30.0\) ms, pulse 2 is sent along the wire from the end at \(x=0 .\) At what position \(x\) do the pulses begin to meet?

Short Answer

Expert verified
The pulses meet at position \( x = 0.31 \mathrm{~m} \).

Step by step solution

01

Calculate Wave Speed

To find the position where the pulses meet, we first need to calculate the speed of the wave traveling along the wire. The formula for the wave speed on a stretched string is \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the wire and \( \mu \) is the linear mass density of the wire. Given \( T = 250 \) N and the mass of the wire is 100 g (0.1 kg), the length of the wire is 10 m, so \( \mu = \frac{0.1 \mathrm{~kg}}{10 \mathrm{~m}} = 0.01 \mathrm{~kg/m} \). Therefore, \( v = \sqrt{\frac{250}{0.01}} = \sqrt{25000} = 158.11 \mathrm{~m/s} \).
02

Calculate the Position of Pulse 1 at First Meeting

Pulse 1 travels from the end at \( x = 10 \mathrm{~m} \) in the negative \( x \) direction. It travels a distance \( x_1 \) such that it meets pulse 2. Let \( t_m \) be the time at which they meet. Thus, \( x_1 = 10 - v \cdot t_m \).
03

Calculate the Position of Pulse 2 at First Meeting

Pulse 2 travels from \( x = 0 \mathrm{~m} \) in the positive \( x \) direction. Given that this pulse is initiated 30 ms after pulse 1, the distance \( x_2 \) it covers by the time of meeting is \( x_2 = v \cdot (t_m - 0.03) \).
04

Set Distances Equal to Find Meeting Point

Since pulse 1 moves towards 0 and pulse 2 moves away from 0, they meet when \( x_1 = x_2 \). Therefore, solve the equation \( 10 - v \cdot t_m = v \cdot (t_m - 0.03) \). Substituting for \( v = 158.11 \mathrm{~m/s} \), we have:\[10 - 158.11t_m = 158.11(t_m - 0.03)\]Simplifying gives:\[10 = 158.11 \times 0.03\]\[t_m = \frac{10}{2 \times 158.11} + 0.03 = 0.0613 \text{ s}\]
05

Calculate x-Position at Meeting Point

Substitute \( t_m = 0.0613 \) s back into either expression for \( x_1 \) or \( x_2 \). Using \( x_1 = 10 - 158.11 \cdot 0.0613 \):\[x_1 = 10 - 9.69 = 0.31 \mathrm{~m}\]Thus, the pulses begin to meet at \( x = 0.31 \mathrm{~m} \).

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Key Concepts

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

Tension in Wire
Understanding the tension in a wire is crucial when dealing with wave motion, especially for calculating wave speed. Tension refers to the force that is applied along the wire, stretching it. It is measured in newtons (N). In our exercise, the wire has a specified tension of 250 N. Tension is a direct force that affects how waves move along the wire.
When tension increases, the force pulling the material tight increases. This means the speed of a wave traveling on the wire also increases. This is because wave speed is related to the tension through the wave speed formula:
  • \( v = \sqrt{\frac{T}{\mu}} \)
In this formula, \( T \) is the tension, and \( \mu \) is the linear mass density. Our example uses this formula to determine how fast the waves move relative to the tension. Therefore, higher tension results in a faster wave due to the stronger pulling force.
Linear Mass Density
Linear mass density (symbol \( \mu \)) represents the mass per unit length of the wire. It is measured in kilograms per meter (kg/m). In the exercise, it's computed as \( \mu = \frac{0.1 \, ext{kg}}{10 \, ext{m}} = 0.01 \, ext{kg/m} \). This measure tells us how much mass is in each meter of the wire, which directly affects how waves travel through the wire.
To calculate wave speed, we use the linear mass density as a factor in the equation \( v = \sqrt{\frac{T}{\mu}} \). The wave speed is inversely proportional to the square root of the linear mass density. This means that for a given tension, increasing the wire's mass density will decrease the wave speed. Understanding this relationship is important when designing systems that use wires to convey mechanical signals, like musical instruments or certain mechanical devices.
Pulse Meeting Point
The exercise involves finding the point where two pulses meet on the wire. The meeting point is where the two traveling pulses, sent from opposite ends, converge. First, we calculate each pulse's travel distance using the speed obtained from the wave speed formula.
When both pulses are traveling toward each other, their meeting point can be found by ensuring their traveled distances add up to the entire length of the wire. For our case:
  • Pulse 1 starts from \( x = 10 \) m.
  • Pulse 2 starts from \( x = 0 \) m.
By comparing their equations, we can solve for the time \( t_m \) when they meet. Substituting this back gives us the position \( x \) on the wire, which mathematically proves to be at \( x = 0.31 \) m from the starting end at \( x = 10 \) m. This is crucial when synchronizing wave motions, such as in communication systems.
Wave Motion
Wave motion in a medium like a wire involves energy being transferred through oscillations. Unlike rigid movement, the particles in the wire oscillate around an equilibrium position, creating waves.
Key points about wave motion include:
  • Waves can be longitudinal or transverse, but in a taut wire, they are usually transverse, meaning the displacement is perpendicular to the direction of wave travel.
  • Wave speed is critical, affected by the medium's physical properties, notably tension and linear mass density.
  • Understanding wave behavior in such setups is crucial for practical applications in engineering and physics.
Waves carry both energy and information, which is why understanding wave motion principles helps in applications ranging from musical instruments to telecommunication technologies.

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Most popular questions from this chapter

The tension in a wire clamped at both ends is doubled without appreciably changing the wire's length between the clamps. What is the ratio of the new to the old wave speed for transverse waves traveling along this wire?

The equation of a transverse wave traveling along a string is $$ y=0.15 \sin (0.79 x-13 t) $$ in which \(x\) and \(y\) are in meters and \(t\) is in seconds. (a) What is the displacement \(y\) at \(x=2.3 \mathrm{~m}, t=0.16 \mathrm{~s} ?\) A second wave is to be added to the first wave to produce standing waves on the string. If the second wave is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t)\), what are (b) \(y_{m},(\mathrm{c})\) \(k\), (d) \(\omega\), and (e) the correct choice of sign in front of \(\omega\) for this second wave? (f) What is the displacement of the resultant standing wave at \(x=2.3 \mathrm{~m}, t=0.16 \mathrm{~s}\) ?

A human wave. During sporting events within large, densely packed stadiums, spectators will send a wave (or pulse) around the stadium (Fig. \(16-29)\). As the wave reaches a group of spectators, they stand with a cheer and then sit. At any instant, the width \(w\) of the wave is the distance from the leading edge (people are just about to stand) to the trailing edge (people have just sat down). Suppose a human wave travels a distance of 853 seats around a stadium in \(39 \mathrm{~s}\), with spectators requiring about \(1.8 \mathrm{~s}\) to respond to the wave's passage by standing and then sitting. What are (a) the wave speed \(v\) (in seats per second) and (b) width \(w\) (in number of seats)?

93 A traveling wave on a string is described by $$ y=2.0 \sin \left[2 \pi\left(\frac{t}{0.40}+\frac{x}{80}\right)\right] $$ where \(x\) and \(y\) are in centimeters and \(t\) is in seconds. (a) For \(t=0\), plot \(y\) as a function of \(x\) for \(0 \leq x \leq 160 \mathrm{~cm} .\) (b) Repeat (a) for \(t=0.05 \mathrm{~s}\) and \(t=0.10 \mathrm{~s}\). From your graphs, determine (c) the wave speed and (d) the direction in which the wave is traveling.

A string along which waves can travel is \(2.70 \mathrm{~m}\) long and has a mass of \(260 \mathrm{~g}\). The tension in the string is \(36.0 \mathrm{~N}\). What must be the frequency of traveling waves of amplitude \(7.70 \mathrm{~mm}\) for the average power to be \(85.0 \mathrm{~W} ?\)

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