Explanation Given data: The following table shows the vibration frequencies of the string and the corresponding number of antinodes. string # Number of antinodes f ( Hz ) 1 1 220 2 2 670 3 1 440 4 3 360 Formula used: The fundamental frequency of the wave is, f n = n ( v 2 l ) Here, f n is the frequency of the wave, v is the speed of the wave, n is the number of antinodes, and l is the length of the medium. The speed of the transverse wave is: v = F μ = F ( m l ) Here, v is the speed of the transverse wave, F is the force exerted, and μ is the mass per unit length. Explanation: The expression for fundamental frequency in rope 1. f 1 = n ( v 1 2 l ) Substitute 1 for n and 220 Hz for f 1 . 220 Hz = 1 ( v 1 2 l ) v 1 = 440 l The expression for the speed of the transverse wave in rope 1. v 1 = F ( m 1 l ) Substitute 440 l for v 1 . 440 l = F ( m 1 l ) Squaring both sides. m 1 = F ( 440 ) 2 l Again, the expression for fundamental frequency in rope 2. f 2 = n ( v 2 2 l ) Substitute 2 for n and 670 Hz for f 2 . 670 Hz = 2 ( v 2 2 l ) v 2 = 670 l The expression for the speed of the transverse wave in rope 2. v 2 = F ( m 2 l ) Substitute 670 l for v 2 . 670 l = F ( m 2 l ) Squaring both sides: m 2 = F ( 670 ) 2 l Again, the expression for fundamental frequency in rope 3

2nd Edition

ISBN: 9780134601823

Author: ETKINA, Eugenia, Planinšič, G. (gorazd), Van Heuvelen, Alan

Publisher: Pearson,

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Chapter 11, Problem 59P

To determine

The rank of the mass of the string from smallest to largest, if four strings of equal lengths are pulled with equal forces. A table for vibration frequency and the corresponding antinodes is provided as:

string #Number ofantinodesf(Hz)11220226703144043360

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Chapter 11, Problem 58P

Chapter 11, Problem 60P

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The rank of the mass of the string from smallest to largest, if four strings of equal lengths are pulled with equal forces. A table for vibration frequency and the corresponding antinodes is provided as: string # Number of antinodes f ( Hz ) 1 1 220 2 2 670 3 1 440 4 3 360

Solution Summary: The author explains how to determine the mass of the string from smallest to largest, if four strings of equal lengths are pulled with equal forces.

Concept explainers

The rank of the mass of the string from smallest to largest, if four strings of equal lengths are pulled with equal forces. A table for vibration frequency and the corresponding antinodes is provided as:

string #Number ofantinodesf(Hz)11220226703144043360

Want to see the full answer?

Chapter 11 Solutions

The rank of the mass of the string from smallest to largest, if four strings of equal lengths are pulled with equal forces. A table for vibration frequency and the corresponding antinodes is provided as: string # Number of antinodes f ( Hz ) 1 1 220 2 2 670 3 1 440 4 3 360

Solution Summary: The author explains how to determine the mass of the string from smallest to largest, if four strings of equal lengths are pulled with equal forces.

Concept explainers

The rank of the mass of the string from smallest to largest, if four strings of equal lengths are pulled with equal forces. A table for vibration frequency and the corresponding antinodes is provided as: string #Number ofantinodesf(Hz)11220226703144043360

Want to see the full answer?

Chapter 11 Solutions

The rank of the mass of the string from smallest to largest, if four strings of equal lengths are pulled with equal forces. A table for vibration frequency and the corresponding antinodes is provided as:

string #Number ofantinodesf(Hz)11220226703144043360