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Repeat Exercise 14.13, but assume that at t = 0 the block

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 12E Chapter 14

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 12E

Repeat Exercise 14.13, but assume that at t = 0 the block has velocity - 4.00 m > s and displacement + 0.200 m. 14.13 . A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

Step-by-Step Solution:

Solution 12E Introduction We have to calculate the amplitude A, phase angle . Then we have to write the equation of motion. Step 1 (a) The total energy at the given time is the kinetic energy of the mass plus the potential energy of the spring. Hence the initial energy of the system is E = k2 + mv 2 2 Now if A is the amplitude then we have 1 2 E = k2 So we have 2kA = k2 + mv 2 2 A = kx +mv k (350 N/m)(0.200 m) +(2.00 kg)(4.00 m/s) = 350 N/m = 0.363 m So the amplitude of the motion is 0.363 m. Step 2 (b) Let us consider that the equation of motion is x = Asin(t + ) Now at t = 0, we have 0.200 m = (0.363 m)sin() sin() = 0.551 Now since the velocity of the mass is given by v = dt = Acos(t + ) Now at t = 0, we have v = 4.00 m/s. Also the angular frequency of the motion is = m = 2.00 kg13.2 rad/s Hence we have ( 4 m/s) = (0.363 m)(13.2 rad/s)cos() cos() = 0.835 So we have tan() = 0.660 1 = tan ( 0.660) = 2.56 rad Hence the phase angle is 2.56 rad.

Step 3 of 3

Chapter 14, Problem 12E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Repeat Exercise 14.13, but assume that at t = 0 the block

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