/*! 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 66 A string of length \(L\) consist... [FREE SOLUTION] | 91Ó°ÊÓ

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A string of length \(L\) consists of two sections. The left half has mass per unit length \(\mu=\mu_{0} / 2,\) while the right has a mass per unit length \(\mu^{\prime}=3 \mu=3 \mu_{0} / 2 .\) Tension in the string is \(T_{0} .\) Notice from the data given that this string has the same total mass as a uniform string of length \(L\) and mass per unit length \(\mu_{0}\). (a) Find the speeds \(v\) and \(v^{\prime}\) at which transverse pulses travel in the two sections. Express the speeds in terms of \(T_{0}\) and \(\mu_{0},\) and also as multiples of the speed \(v_{0}=\left(T_{0} / \mu_{0}\right)^{1 / 2} .\) (b) Find the time interval required for a pulse to travel from one end of the string to the other. Give your result as a multiple of \(\Delta t_{0}=L / v_{0}\).

Short Answer

Expert verified
The speeds of the transverse pulses on the left and right sections of the string are \(2v_{0}\) and \(v_{0}/\sqrt{3}\) respectively, and the time interval required for a pulse to travel from one end of the string to the other is \((1/4+\sqrt{3}/2)(\Delta t_{0})\).

Step by step solution

01

Determine Speeds of two Sections

First, we apply the wave speed formula \(v=\sqrt{T/\mu}\) where \(T\) is the tension (which is given as \(T_{0}\)) and \(\mu\) is the mass per unit length. For the left half of the string, \(\mu=\mu_{0}/2\) and for the right half, \(\mu'=3\mu=3\mu_{0}/2\). Substituting these values in the wave speed formula, we get: \(v=\sqrt{T_{0}/(\mu_{0}/2)}\) and \(v'=\sqrt{T_{0}/(3\mu_{0}/2)}\).
02

Express Speeds in terms of \(v_{0}\)

We have the relation \(v_{0}=\sqrt{T_{0}/\mu_{0}}\). Getting a common denominator and square rooting to express \(v\) and \(v'\) as multiples of \(v_{0}\), we have: \(v=2v_{0}\) and \(v'=v_{0}/\sqrt{3}\).
03

Find Time Taken for a Pulse to Travel Across String

The time taken is simply the distance (\(L/2\) for each section) divided by speed (\(v\) and \(v'\) respectively). So: \(\Delta t=(L/2v)+(L/2v')=(L/2(2v_{0}))+((L/2)/(v_{0}/\sqrt{3}))\)=\(L/4v_{0}+(L\sqrt{3}/2v_{0})=(1/4+\sqrt{3}/2)(\Delta t_{0})\) where \( \Delta t_{0}=L/v_{0}\).

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Key Concepts

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

Transverse Pulses
Imagine you flick one end of a rope and watch a bump travel along it; this bump is what we call a transverse pulse. In physics, a transverse pulse is a disturbance that moves perpendicular to the direction of the wave. When you create this pulse on a string, you're setting off a wave that travels from one point to another, transporting energy but not matter along the medium— in this case, the string.

These pulses move according to certain principles of wave mechanics, one of which is the wave speed formula. The speed at which they travel depends on a variety of factors, including the tension in the string and its mass per unit length. To visualize this, if you tighten the rope (increase the tension) and send another pulse, you'll notice it travels faster; this is because a greater tension results in a quicker pulse movement.
Mass per Unit Length
Mass per unit length, often symbolized as \(\mu\), is a critical factor that influences the speed of transverse waves on a string or rope. It's defined as the mass of the string divided by its total length. You can think of it as how 'heavy' the string feels per unit of length. The heavier a string is per unit length, the slower a transverse pulse will travel through it.

When solving problems regarding wave propagation, understanding the role of \(\mu\) is essential. In simpler terms, if we had two identical strings, but one was made of a heavier material, waves would travel slower on the heavier one due to its greater mass per unit length. This concept is perfectly demonstrated in the exercise, where the string is divided into two sections with different values of \(\mu\), resulting in different wave speeds.
Tension in a String
The tension in a string, which we'll denote with \(T\), also plays a vital part in the behavior of transverse pulses. Tension is the force exerted along the string's length, typically caused by forces pulling on either end of the string. It's crucial to understand that the tension in a string affects the speed of wave propagation: the greater the tension, the faster the wave speed.

For a given mass per unit length, a string with more tension will allow transverse waves to travel faster compared to one with less tension. That's because higher tension forces make the string more resistant to the bending caused by the pulse, enabling the pulse to zip along with less resistance. Our exercise demonstrates this when expressing the speed of a pulse as a function of the tension in the string, \(T_{0}\), giving us a clearer picture of wave mechanics involved.
Wave Propagation
Wave propagation refers to the way waves travel through a medium, which in the case of our exercise is a string. This covers the journey of energy from one end of the medium to the other, often without the transport of matter. Transverse pulses on a string are a clear example of wave propagation, where the pulse's movement from one end to the other demonstrates how energy is transferred through the string.

Understanding wave propagation is crucial for solving problems involving waves, as it helps to determine how fast the energy will move across a distance. The exercise looks at this by finding the time it takes for a pulse to traverse the string's length. By doing this, you not only grasp the abstract theory behind waves but also apply it to practical scenarios, enhancing both your conceptual understanding and problem-solving skills.

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Most popular questions from this chapter

A sinusoidal wave of wavelength \(2.00 \mathrm{m}\) and amplitude \(0.100 \mathrm{m}\) travels on a string with a speed of \(1.00 \mathrm{m} / \mathrm{s}\) to the right. Initially, the left end of the string is at the origin. Find (a) the frequency and angular frequency, (b) the angular wave number, and (c) the wave function for this wave. Determine the equation of motion for (d) the left end of the string and (e) the point on the string at \(x=\) \(1.50 \mathrm{m}\) to the right of the left end. (f) What is the maximum speed of any point on the string?

The wave function for a traveling wave on a taut string is (in SI units) $$y(x, t)=(0.350 \mathrm{m}) \sin (10 \pi t-3 \pi x+\pi / 4)$$ (a) What are the speed and direction of travel of the wave? (b) What is the vertical position of an element of the string at \(t=0, x=0.100 \mathrm{m} ?\) (c) What are the wavelength and frequency of the wave? (d) What is the maximum magnitude of the transverse speed of the string?

A transverse sinusoidal wave on a string has a period \(T=25.0 \mathrm{ms}\) and travels in the negative \(x\) direction with a speed of \(30.0 \mathrm{m} / \mathrm{s} .\) At \(t=0,\) a particle on the string at \(x=0\) has a transverse position of \(2.00 \mathrm{cm}\) and is traveling downward with a speed of \(2.00 \mathrm{m} / \mathrm{s} .\) (a) What is the amplitude of the wave? (b) What is the initial phase angle? (c) What is the maximum transverse speed of the string? (d) Write the wave function for the wave.

Two points \(A\) and \(B\) on the surface of the Earth are at the same longitude and \(60.0^{\circ}\) apart in latitude. Suppose that an earthquake at point \(A\) creates a \(P\) wave that reaches point \(B\) by traveling straight through the body of the Earth at a constant speed of \(7.80 \mathrm{km} / \mathrm{s} .\) The earthquake also radiates a Rayleigh wave, which travels across the surface of the Earth in an analogous way to a surface wave on water, at \(4.50 \mathrm{km} / \mathrm{s}\). (a) Which of these two seismic waves arrives at \(B\) first? (b) What is the time difference between the arrivals of the two waves at \(B ?\) Take the radius of the Earth to be \(6370 \mathrm{km}\).

For a certain transverse wave, the distance between two successive crests is \(1.20 \mathrm{m},\) and eight crests pass a given point along the direction of travel every 12.0 s. Calculate the wave speed.

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