Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Question
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Chapter 11, Problem 45P
To determine

To Find: The amplitude of the object.

Expert Solution & Answer
Check Mark

Answer to Problem 45P

  0.99 m

Explanation of Solution

Given info:

Frequency  f = 0.5 Hz

Formula used:

The expression for acceleration.

  amax = A×ω2

  amax is greater than the acceleration due to gravity.

Thus,

  A×ω2 > gA > gω2

Here,

  g is acceleration due to gravity.

  A is amplitude of oscillation.

  ω is angular frequency.

Calculation:

Assume the earthquake wave moving the ground vertically since it is a transverse wave. Object sitting on the ground would be moving with some harmonic motion. The object has normal force and self weight.

If object losses contact with the ground the normal force of an object is equal to zero. Only force acting on the object is self weight. Substitute the given values:

  A > gω2 = g4×π2×f2As,ω = 2πf

Limiting factor of amplitude, A=9.84×π2×0.520.99 m

Conclusion:

Thus, the amplitude in such a way that the object begins to leave contact with the ground is 0.99 m .

Chapter 11 Solutions

Physics: Principles with Applications

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