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The tension in a wire clamped at both ends is halved 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?

Short Answer

Expert verified
The ratio of the new wave speed to the old wave speed is \( \frac{1}{\sqrt{2}} \).

Step by step solution

01

Understanding Wave Speed Formula

The speed of a wave on a stretched string depends on the tension in the string and the linear density of the string. The formula for wave speed \( v \) is given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string and \( \mu \) is the linear mass density.
02

Calculate Old Wave Speed

Let the initial tension be \( T_1 \) and the linear mass density \( \mu \). Then, the initial wave speed \( v_1 \) is given by \( v_1 = \sqrt{\frac{T_1}{\mu}} \).
03

Calculate New Wave Speed

When the tension is halved, the new tension \( T_2 \) will be \( T_2 = \frac{T_1}{2} \). The new wave speed \( v_2 \) is \( v_2 = \sqrt{\frac{T_2}{\mu}} = \sqrt{\frac{T_1/2}{\mu}} = \sqrt{\frac{T_1}{2\mu}} = \frac{1}{\sqrt{2}} \cdot \sqrt{\frac{T_1}{\mu}} \).
04

Ratio of New to Old Wave Speed

We seek \( \frac{v_2}{v_1} \). Using the expressions from previous steps: \( \frac{v_2}{v_1} = \frac{\frac{1}{\sqrt{2}} \cdot \sqrt{\frac{T_1}{\mu}}}{\sqrt{\frac{T_1}{\mu}}} = \frac{1}{\sqrt{2}} \). Thus, the ratio \( \frac{v_2}{v_1} = \frac{1}{\sqrt{2}} \).

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Key Concepts

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

Tension in String
Tension in a string plays a crucial role in determining the speed of waves traveling through it. When we talk about tension, we refer to the force stretching the string, usually exerted by external forces at the ends of the string. This force is measured in Newtons and significantly impacts how fast the wave moves.
In the exercise, the tension in the string was halved, which translates to a decrease in the stretching force. With less tension, the wave does not travel as quickly through the string, which affects the dynamics of wave propagation. In terms of physics principles, reducing tension tends to make the string "looser," thereby allowing waves to move with less acceleration compared to a tightly stretched string.
This concept helps us understand why the wave speed changes when tension is altered.
Linear Mass Density
Linear mass density, denoted as \( \mu \), is another critical factor influencing wave speed in strings. It represents the mass of the string per unit length and is usually expressed in kilograms per meter. A higher linear mass density means more mass is present along the same length of string, which affects how energy is transmitted through the string.
In our scenario, the linear mass density is assumed constant. This means the mass per unit length does not change, even when tension is altered. It highlights the fact that while tension affects the wave speed, the inherent characteristics of the string, encapsulated by its linear mass density, remain the same.
Understanding linear mass density is important because it allows us to see how different factors can independently impact wave speed, even if the material of the string remains unchanged.
Transverse Waves
Transverse waves are a type of wave where the motion of the medium is perpendicular to the direction of the wave propagation. When you pluck a guitar string, the vibrations that create sound travel in the form of transverse waves. In this exercise, the waves moving along the wire are transverse since their oscillations occur at right angles to the length of the string.
This concept is essential to explore as it helps us visually understand the interactions involved when waves travel along a string. Unlike longitudinal waves, which move in the direction of wave travel (like sound waves in air), transverse waves wiggle side-to-side as they propagate. This difference impacts how energy is transferred along the medium and how wave speed calculations are made.
Wave Speed Formula
The wave speed formula, \( v = \sqrt{\frac{T}{\mu}} \), is fundamental to understanding how various factors influence the speed at which waves travel along a string. Here, \( v \) stands for wave speed, \( T \) for tension, and \( \mu \) for linear mass density.
By applying this formula, as in the exercise, we see how a change in tension affects wave speed. Initially, the tension \( T_1 \) produces a specific wave speed. When the tension is halved to \( T_2 = \frac{T_1}{2} \), the new wave speed becomes \( v_2 = \frac{1}{\sqrt{2}} \cdot v_1 \).
This formula is key because it elegantly ties together physical properties such as tension and mass density, allowing us to predict changes in wave speed when these factors vary. It's an essential tool in both theoretical physics and practical applications, such as tuning musical instruments or designing materials for optimal wave propagation.

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Most popular questions from this chapter

Two waves are generated on a string of length \(4.0 \mathrm{~m}\) to produce a three-loop standing wave with an amplitude of \(1.0 \mathrm{~cm}\). The wave speed is \(100 \mathrm{~m} / \mathrm{s}\). Let the equation for one of the waves be of the form \(y(x, t)=y_{m} \sin (k x+\omega t)\). In the equation for the other wave, what are (a) \(y_{m}\), (b) \(k\), (c) \(\omega\), and (d) the sign in front of \(\omega\) ?

A string oscillates according to the equation $$ y^{\prime}=(0.80 \mathrm{~cm}) \sin \left[\left(\frac{\pi}{3} \mathrm{~cm}^{-1}\right) x\right] \cos \left[\left(40 \pi \mathrm{s}^{-1}\right) t\right] $$ What are the (a) amplitude and (b) speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation? (c) What is the distance between nodes? (d) What is the transverse speed of a particle of the string at the position \(x=2.1 \mathrm{~cm}\) when \(t=0.50 \mathrm{~s} ?\)

A generator at one end of a very long string creates a wave given by $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x+\left(6.00 \mathrm{~s}^{-1}\right) t\right] $$ and a generator at the other end creates the wave $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x-\left(6.00 \mathrm{~s}^{-1}\right) t\right] $$ Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave. For \(x \geq 0\), what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of \(x\) ? For \(x \geq 0\), what is the location of the antinode having the \((\mathrm{g})\) smallest, (h) second smallest, and (i) third smallest value of \(x\) ?

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 \(51 \mathrm{~s}\), 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)?

The heaviest and lightest strings on a certain violin have linear densities of \(3.2\) and \(0.26 \mathrm{~g} / \mathrm{m}\). What is the ratio of the diameter of the heaviest string to that of the lightest string, assuming that the strings are of the same material?

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