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Chapter 15: Problem 13

(II) Two children are sending signals along a cord of total mass 0.50 kg tied between tin cans with a tension of \(35 \mathrm{~N}\). It takes the vibrations in the string \(0.50 \mathrm{~s}\) to go from one child to the other. How far apart are the children?

Short Answer

Expert verified
The children are 17.5 meters apart.

Step by step solution

01

Identify the Known Values

We know the total mass of the cord is 0.50 kg and the tension in the cord is \(35\, \text{N}\). It takes vibrations in the string \(0.50\, \text{s}\) to travel from one child to the other.
02

Calculate the Mass per Unit Length

To find the mass per unit length, \( \mu \), of the cord, we will use the formula \( \mu = \frac{m}{L} \), where \(m\) is the total mass of the cord and \(L\) is the length, which is unknown at this point. However, we鈥檒l use \( \mu \) later on to relate other quantities.
03

Use the Wave Speed Formula

The wave speed \( v \) on a string is given by \( v = \sqrt{\frac{T}{\mu}} \), where \(T\) is the tension (35 N) and \(\mu\) is the mass per unit length. We need to use this relationship to eventually find the distance \(L\).
04

Relate Time, Speed, and Distance

The speed of the wave \( v \) is equal to \( \frac{L}{t} \), where \(L\) is the length of the cord (which is the distance between the children) and \(t\) is the time taken (0.50 s). Thus, \( L = v \times t \).
05

Solve for \(L\)

First, express \(\mu\): \(\mu = \frac{0.50}{L} \). Then, substituting in the wave speed equation, \(v = \sqrt{\frac{35}{\frac{0.50}{L}}}\). Simplifying gives \(v = \sqrt{70L}\). The travel time equation \(L = v \times 0.50\) becomes \(L = \sqrt{70L} \times 0.50\). Squaring both sides: \(L^2 = 70L \times 0.25\), leading to \(L^2 = 17.5L\) and \(L = 17.5\).
06

Final Calculation and Answer

With \(L = 17.5\), the distance between the children is found to be 17.5 meters.

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Key Concepts

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

Tension in a String
Understanding tension in a string is crucial to analyzing problems involving waves traveling through a string or cord. Tension refers to the force exerted along the string or rope, often caused by it being pulled tight. In the context of our exercise, the string is under a tension of 35 N. Tension affects how a wave travels through a string. As tension increases, the speed of the wave also increases. This occurs because the force pulls the particles in the string closer together, making it easier for waves to propagate. To visualize tension, think about pulling both ends of a rope. The rope becomes tighter and if you pluck it, a wave travels more swiftly compared to when the rope is loosely held. This principle holds true for any wave on a string, such as the sound waves produced by a guitar string being strummed.
Mass per Unit Length
The mass per unit length, often denoted as \( \mu \), is a measure of how much mass the string has per length unit. It is calculated using the formula \( \mu = \frac{m}{L} \), where \( m \) is the total mass and \( L \) is the total length of the string.This value is key in determining the wave speed in a string. With more mass per unit length, waves travel slower since it takes more effort to move the heavier sections of the string. In our example, though the total cord mass is given, the length is initially unknown. Despite this, understanding \( \mu \) is necessary as it directly influences the wave speed formula: \( v = \sqrt{\frac{T}{\mu}} \). This equation shows a direct relationship: lower \( \mu \) results in faster wave speed, given constant tension.
Distance Calculation
Distance calculation in this problem involves determining how far a wave travels in a given time. The relationship between wave speed, time, and distance is essential here.Wave speed \( v \) is calculated using tension \( T \) and mass per unit length \( \mu \). Once \( v \) is determined, it relates to distance through the formula \( L = v \times t \) where \( t \) is the time taken for the wave to travel the distance.In our exercise, vibrations in the string take 0.50 seconds to travel between the two children. By substituting the calculated \( v \) into the formula, we find the distance \( L \). Through precise calculations, we determine that the distance is 17.5 meters, revealing how understanding and manipulating these relationships allows us to find the sought after separation between the children.

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Most popular questions from this chapter

Two strings on a musical instrument are tuned to play at \(392 \mathrm{~Hz}(\mathrm{G})\) and \(494 \mathrm{~Hz}(\mathrm{~B}),(a)\) What are the frequencies of the first two overtones for each string? \((b)\) If the two strings have the same length and are under the same tension, what must be the ratio of their masses \(\left(m_{\mathrm{G}} / m_{\mathrm{A}}\right) ?(c)\) If the strings, instead, have the same mass per unit length and are under the same tension, what is the ratio of their lengths \(\left(\ell_{\mathrm{G}} / \ell_{\mathrm{A}}\right) ?(d)\) If their masses and lengths are the same, what must be the ratio of the tensions in the two strings?

(1) In an earthquake, it is noted that a footbridge oscillated up and down in a one-loop (fundamental standing wave) pattern once every 1.5 \(\mathrm{s}\) . What other possible resonant periods of motion are there for this bridge? What frequencies do they correspond to?

(II) The displacement of a standing wave on a string is given by \(D=2.4 \sin (0.60 x) \cos (42 t),\) where \(x\) and \(D\) are in centimeters and \(t\) is in seconds. (a) What is the distance (cm) between nodes? (b) Give the amplitude, frequency, and speed of each of the component waves. (c) Find the speed of a particle of the string at \(x=3.20 \mathrm{~cm}\) when \(t=2.5 \mathrm{~s}\).

An earthquake-produced surface wave can be approximated by a sinusoidal transverse wave. Assuming a frequency of \(0.60 \mathrm{~Hz}\) (typical of earthquakes, which actually include a mixture of frequencies), what amplitude is needed so that objects begin to leave contact with the ground? [Hint: Set the acceleration \(a>g .\)

(II) A cord of length \(1.0 \mathrm{~m}\) has two equal-length sections with linear densities of \(0.50 \mathrm{~kg} / \mathrm{m}\) and \(1.00 \mathrm{~kg} / \mathrm{m}\). The tension in the entire cord is constant. The ends of the cord are oscillated so that a standing wave is set up in the cord with a single node where the two sections meet. What is the ratio of the oscillatory frequencies?
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