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You wish to decrease the speed of a wave traveling on a string to half its current value by changing the tension in the string. By what factor must you decrease the tension in the string? a) 1 c) 2 e) none of the above b) \(\sqrt{2}\) d) 4

Short Answer

Expert verified
Answer: (d) 4

Step by step solution

01

Write the given information

The current wave speed is \(v\) and we wish to decrease it to half its current value, which is \(\frac{1}{2}v\). The goal is to find the factor x by which the tension must be decreased, i.e., the new tension should be \(T' = \frac{T}{x}\).
02

Write the wave speed formula for new tension

Using the formula \(v = \sqrt{\frac{T}{\mu}}\), we can write the new wave speed formula for the decreased tension as follows: \(\frac{1}{2}v = \sqrt{\frac{T'}{\mu}} = \sqrt{\frac{\frac{T}{x}}{\mu}}\)
03

Solve for the factor x

Now, we need to solve the equation above for the factor x. First, we can square both sides to get rid of the square root: \((\frac{1}{2}v)^2 = (\sqrt{\frac{\frac{T}{x}}{\mu}})^2\) \(\frac{1}{4}v^2 = \frac{T}{x\mu}\) Next, we use the initial wave speed formula \(v = \sqrt{\frac{T}{\mu}}\) and rearrange it to express \(T\) in terms of \(v\) and \(\mu\): \(T = v^2\mu\) Substitute this expression into the equation above: \(\frac{1}{4}v^2 = \frac{v^2\mu}{x\mu}\) Now, we can cancel \(v^2\) and \(\mu\) from both sides and solve for x: \(x = 4\)
04

Find the correct answer in the given options

The factor by which we must decrease the tension in the string is 4. Therefore, the correct answer is option (d).

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Key Concepts

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

Wave Speed Formula
Understanding the wave speed formula is essential when dealing with the motion of waves on a string. The formula itself is expressed as \( v = \sqrt{\frac{T}{\mu}} \), where \( v \) represents the speed of the wave, \( T \) is the tension in the string, and \( \mu \) is the mass per unit length of the string.

From this fundamental relationship, it's clear that wave speed is directly related to the tension applied: the greater the tension, the faster the wave travels. Conversely, if the tension decreases, the speed of the wave also decreases. This principle helps us understand the impact of tension adjustments on wave speed. By modifying the tension, one can control the speed of waves, which is a practical aspect in various technical and musical applications.

When the exercise asks for the wave speed to be reduced to half its current value, it effectively requires us to alter the tension based on this foundational formula. A solid grasp of this relationship between tension and wave speed allows us to manipulate the conditions of wave propagation to our advantage, or in this case, to meet the specific exercise requirement.
Tension in string
Tension plays a crucial role in the behavior of waves on a string. This is because tension is the force that provides the restoring force necessary for the wave to propagate. In the context of our problem, we're investigating how altering this force affects the wave's speed.

Understanding Tension

Tension can be thought of as a pulling force that keeps the string taut, enabling waves to move along its length. It's an intrinsic part of wave dynamics on a string, and it’s important to realize that tension is a scalar quantity that only has magnitude, and no direction.

When it comes to changing the wave's speed, the exercise prompts us to adjust the tension by a specific factor. Since wave speed on a string is proportional to the square root of the tension, reducing the speed to half its original value implies a significant reduction in tension, specifically to one-fourth. This outcome is the square of the speed reduction factor since we're dealing with the square root relationship.
Wave Mechanics
Wave mechanics is the branch of physics that deals with the study of waves. Waves on a string are a common example of mechanical waves that require a medium (in this case, the string) to travel through.

Wave Behavior on Strings

Several factors affect the mechanics of a wave on a string, including tension, mass per unit length, and the way the wave is generated. For the purpose of the exercise, we're primarily concerned with the effect of tension on the wave's propagation speed.

The wave mechanics on a string are governed by principles found in the wave speed formula. Whenever the tension is altered, the speed of the wave responds accordingly. In this educational context, the manipulation of wave speed by adjusting tension offers a tangible demonstration of wave mechanics principles. The exercise illustrates the direct relationship between tension and speed, emphasizing the predictive nature of physical laws in wave behavior.

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Most popular questions from this chapter

A rubber band of mass \(0.4245 \mathrm{~g}\) is stretched between two fingers. The overall stretched length of the band is \(20.91 \mathrm{~cm} .\) One side of the band is plucked, setting up a vibration in \(8.117 \mathrm{~cm}\) of the band's stretched length. The lowest-frequency vibration that can be set up on this part of the rubber band is \(184.2 \mathrm{~Hz}\). What is the tension in each side of the rubber band? Assume that the band stretches uniformly.

A string with linear mass density \(\mu=0.0250 \mathrm{~kg} / \mathrm{m}\) under a tension of \(F=250 . \mathrm{N}\) is oriented in the \(x\) -direction. Two transverse waves of equal amplitude and with a phase angle of zero (at \(t=0\) ) but with different angular frequencies \((\omega=3000 . \mathrm{rad} / \mathrm{s}\) and \(\omega / 3=1000 . \mathrm{rad} / \mathrm{s})\) are created in the string by an oscillator located at \(x=0 .\) The resulting waves, which travel in the positive \(x\) -direction, are reflected at a distant point where both waves have a node, so there is a similar pair of waves traveling in the negative \(x\) -direction. Find the values of \(x\) at which the first two nodes in the standing wave are produced by these four waves.

Two waves traveling in opposite directions along a string fixed at both ends create a standing wave described by \(y(x, t)=\) \(1.00 \cdot 10^{-2} \sin (25.0 x) \cos (1200 . t) .\) The string has a linear mass density of \(0.0100 \mathrm{~kg} / \mathrm{m},\) and the tension in the string is supplied by a mass hanging from one end. If the string vibrates in its third harmonic, calculate (a) the length of the string, (b) the velocity of the waves, and (c) the mass of the hanging mass.

Consider a wave on a string, with amplitude \(A\) and wavelength \(\lambda\), traveling in one direction. Find the relationship between the maximum speed of any portion of string, \(v_{\text {max }}\), and the wave speed, \(v\)

Bob is talking to Alice using a tin can telephone, which consists of two steel cans connected by a 20.0 - m-long taut steel wire (see the figure). The wire has a linear mass density of \(6.13 \mathrm{~g} / \mathrm{m},\) and the tension on the wire is \(25.0 \mathrm{~N}\). The sound waves leave Bob's mouth, are collected by the can on the left, and then create vibrations in the wire, which travel to Alice's can and are transformed back into sound waves in air. Alice hears both the sound waves that have traveled through the wire (wave 1 ) and those that have traveled through the air (wave 2), bypassing the wire. Do these two kinds of waves reach her at the same time? If not, which wave arrives sooner and by how much? The speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\). Assume that the waves on the string are transverse.

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