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A string has its tension doubled; all else kept constant, what happens to the speed of transverse waves that can be set up on the string?

Short Answer

Expert verified
The speed increases by a factor of \( \sqrt{2} \).

Step by step solution

01

Understanding the Formula

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

Identifying Given Changes

In this problem, the tension \( T \) is doubled, meaning that the new tension is \( 2T \). The linear mass density \( \mu \) remains constant.
03

Substituting New Tension

Substitute the new tension into the formula: \( v' = \sqrt{\frac{2T}{\mu}} \).
04

Simplifying the Expression

The expression \( \sqrt{\frac{2T}{\mu}} \) can be rewritten as \( \sqrt{2} \times \sqrt{\frac{T}{\mu}} \). This tells us that \( v' = \sqrt{2} \times v \).
05

Conclude New Speed

Since \( v' = \sqrt{2} \times v \), the speed of the transverse wave increases by a factor of \( \sqrt{2} \) when the tension is doubled, assuming all other factors remain constant.

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Key Concepts

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

Transverse Waves
Transverse waves are a fascinating type of wave where the movement of the medium is perpendicular to the direction of the wave's travel. Imagine a string being plucked; the wave travels along the length of the string, but the particles in the string move up and down. This perpendicular motion differentiates them from longitudinal waves, where the movement is parallel.
Transverse waves can be observed in various scenarios like light waves and water waves. However, when discussing them in terms of a string, certain factors influence their behavior. The speed of these waves is notably determined by the tension of the string and its mass per unit length. Understanding how these factors affect wave speed allows us to predict and manipulate the way waves behave across different mediums.
String Tension
String tension plays a crucial role in determining the speed of transverse waves. Tension is the force applied along the string, pulling it tighter. Increasing the tension generally leads to faster wave speeds. This is because with more tension, waves encounter less resistance and can travel further distances more quickly.
In musical instruments, players often adjust string tension to tune the instrument, achieving desired pitch levels based on wave frequency, which is directly related to their speed. In the case of our exercise, by doubling the tension in a string, we effectively increase the wave speed by a factor of \( \sqrt{2} \). This relationship is beautifully captured by the formula \( v = \sqrt{\frac{T}{\mu}} \), where doubling the tension makes the wave travel faster, exemplifying the direct link between tension and wave dynamics.
Physics Formulas
Physics formulas help us define and predict the behaviors of physical phenomena. In studying waves on a string, formula \( v = \sqrt{\frac{T}{\mu}} \) is essential. This formula indicates that the speed \( v \) of a transverse wave is determined by the string's tension \( T \) and its linear mass density \( \mu \).
Understanding this equation is crucial because it bridges theoretical concepts with practical implications. The formula implies that for a fixed mass density, increasing the tension results in faster waves. This can be particularly useful in engineering and music; engineers might use this principle to design materials that efficiently handle vibrations, and musicians harness the same concept to control sound production.
Utilizing such formulas, students can solve complex problems involving waves, predictions of changes in wave speed based on physical manipulations like tension adjustments are made straightforward and easy.

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Most popular questions from this chapter

A string has both its total mass and length doubled; all else kept constant, what happens to the speed of transverse waves that can be set up on the string?

A uniform flexible cable is \(20 \mathrm{~m}\) long and has a mass of \(5.0\) kg. It hangs vertically under its own weight and is vibrated (perpendicularly) from its upper end with a frequency of \(7.0 \mathrm{~Hz}\). (a) Find the speed of a transverse wave on the cable at its midpoint. (b) What are the frequency and wavelength at the midpoint? (b) Because wave crests do not pile up along a string or cable, the number passing one point must be the same as that for any other point. Therefore, the frequency, \(7.0 \mathrm{~Hz}\), is the same at all points. To find the wavelength at the midpoint, we must use the speed we found for that point, \(9.9 \mathrm{~m} / \mathrm{s}\). That gives us $$ \lambda=\frac{v}{f}=\frac{9.9 \mathrm{~m} / \mathrm{s}}{7.0 \mathrm{~Hz}}=1.4 \mathrm{~m} $$

A rod \(120 \mathrm{~cm}\) long is clamped at the center and is stroked in such a way as to give its first overtone. Make a drawing showing the location of the nodes and antinodes, and determine at what other points the rod might be clamped and still emit the same tone.

A flexible cable, \(30 \mathrm{~m}\) long and weighing \(70 \mathrm{~N}\), is stretched by a force of \(2.0 \mathrm{kN}\). If the cable is struck sideways at one end, how long will it take the transverse wave to travel to the other end and return?

A violin string resonates at its fundamental frequency of \(196 \mathrm{~Hz}\). Where along the string must you place your finger so its fundamental becomes \(440 \mathrm{~Hz}\) ? For the fundamental, \(L=\frac{1}{2} \lambda\). Since \(\lambda=\mathrm{v} / f\), it follows that \(f_{1}=\) \(\mathrm{u} / 2 \mathrm{~L}\). Originally, the string of length \(L_{1}\) resonated at a frequency of \(196 \mathrm{~Hz}\), and therefore $$ 196 \mathrm{~Hz}=\frac{v}{2 L_{1}} $$ with a resonance at \(440 \mathrm{~Hz}\), $$ 440 \mathrm{~Hz}=\frac{v}{2 L_{2}} $$ Eliminate \(u\) from these two simultaneous equations and find $$ \frac{L_{2}}{L_{1}}=\frac{196 \mathrm{~Hz}}{440 \mathrm{~Hz}}=0.445 $$ To obtain the desired resonance, the finger must shorten the string to \(0.445\) of its original length.

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