(a) Explanation The vibration in a string moves in the form of sine or cosine wave with certain amplitude. The wave on a string is described by, y ( x , t ) = A cos ( k x − ω t ) (I) Differentiate the above equation with respect to t to find v y . v y = d y d t = − A ω sin ( k x − ω t ) (II) Differentiate the above equation again with respect to t , to find the acceleration a y . a y = d 2 y d t 2 = − A ω 2 cos ( k x − ω t ) (III) At t = 0 , the above equations becomes, y ( x , t ) = A cos ( k x ) v y = − A ω sin ( k x ) a y = − A ω 2 cos ( k x ) Data for the further calculation is, SN x v y a y 1 0 0 − A ω 2 2 π / 4 k − 0 (b) To determine To describe: Whether the particle is moving, speeding up, slowing down or instantaneously not accelerating.

University Physics with Modern Physics (14th Edition)

14th Edition

ISBN: 9780321973610

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Chapter 15, Problem 15.14E

(a)

To determine

To draw: The graph of y , vy and ay as a function of x for the given wave on a string.

(b)

To determine

To describe: Whether the particle is moving, speeding up, slowing down or instantaneously not accelerating.

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Chapter 15, Problem 15.13E

Chapter 15, Problem 15.15E

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Chapter 15 Solutions

University Physics with Modern Physics (14th Edition)

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To draw: The graph of y , v y and a y as a function of x for the given wave on a string.

Solution Summary: The author explains how the vibration in a string moves in the form of sine or cosine wave with certain amplitude.

Concept explainers

To draw: The graph of y , vy and ay as a function of x for the given wave on a string.

To describe: Whether the particle is moving, speeding up, slowing down or instantaneously not accelerating.

Want to see the full answer?

Chapter 15 Solutions