Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 11, Problem 13P

(a)

To determine

To find: The amplitude of the motion.

(a)

Expert Solution
Check Mark

Answer to Problem 13P

The amplitude of the motion is 6.034×10-2 m .

Explanation of Solution

Given:

Mass, m =3 kg

Spring constant, k= 280 N/m

Distance, x = 0.020 m

Velocity, v = 0.55 m/s

Formula used:

The total energy of the spring is the addition of the kinetic energy and potential energy of the spring

  Etotal = K.ESpring + P.ESpringEtotal = 12×m×v2 + 12×k×x2 = 12×k×A2

Here, m is the mass,

  v is the speed of the motion,

  x is the distance moved by the spring,

  k is the spring constant,

  A is the amplitude of the motion.

Calculation:

From the above formula, write the expression for amplitude A.

  A2 = x2 + mk×v2A2 = (0.02)2 + (3280×(0.55)2)A2 = 3.64×10-3 m2A = 6.034×10-2 m

Conclusion:

Thus, the amplitude of the motion is 6.034×10-2 m .

(b)

To determine

To Find: The maximum velocity attained by the object.

(b)

Expert Solution
Check Mark

Answer to Problem 13P

b0.58 m/s

Explanation of Solution

Given:

Mass, m =3 kg

Spring constant, k= 280 N/m

Amplitude, A = 6.034×10-2 m

Formula used:

Energy is called kinetic energy when the object has the maximum velocity, maximum velocity is calculated by the following formula,

  Etotal = 12×k×A2=12×m×v2max

Here, m is the mass,

  vmax is the maximum speed of the motion,

  x is the distance moved by the spring,

  k is the spring constant,

  A is the amplitude of the motion.

Calculation:

Substitute the given values in the above equation to get maximum velocity by following way

  vmax2=k×A2mvmax2 = 280 ×(6.034×10-2)23vmax = 0.58 m/s

Conclusion:

Thus, the maximum velocity attained by the object is 0.58 m/s .

Chapter 11 Solutions

Physics: Principles with Applications

Ch. 11 - Since the density of air decreases with an...Ch. 11 - How did geophysicists determine that part of the...Ch. 11 - Prob. 13QCh. 11 - Prob. 14QCh. 11 - Prob. 15QCh. 11 - Prob. 16QCh. 11 - Prob. 17QCh. 11 - Prob. 18QCh. 11 - Why do the strings used for the lowest-frequency...Ch. 11 - Prob. 20QCh. 11 - Prob. 21QCh. 11 - Prob. 22QCh. 11 - Prob. 1PCh. 11 - Prob. 2PCh. 11 - Prob. 3PCh. 11 - Prob. 4PCh. 11 - Prob. 5PCh. 11 - Prob. 6PCh. 11 - Prob. 7PCh. 11 - Prob. 8PCh. 11 - Prob. 9PCh. 11 - Prob. 10PCh. 11 - Prob. 11PCh. 11 - Prob. 12PCh. 11 - Prob. 13PCh. 11 - Prob. 14PCh. 11 - Prob. 15PCh. 11 - Prob. 16PCh. 11 - Prob. 17PCh. 11 - Prob. 18PCh. 11 - Prob. 19PCh. 11 - Prob. 20PCh. 11 - Prob. 21PCh. 11 - Prob. 22PCh. 11 - Prob. 23PCh. 11 - Prob. 24PCh. 11 - Prob. 25PCh. 11 - Prob. 26PCh. 11 - Prob. 27PCh. 11 - Prob. 28PCh. 11 - Prob. 29PCh. 11 - Prob. 30PCh. 11 - Prob. 31PCh. 11 - Prob. 32PCh. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Prob. 36PCh. 11 - Prob. 37PCh. 11 - Prob. 38PCh. 11 - Prob. 39PCh. 11 - Prob. 40PCh. 11 - Prob. 41PCh. 11 - Prob. 42PCh. 11 - Prob. 43PCh. 11 - Prob. 44PCh. 11 - Prob. 45PCh. 11 - Prob. 46PCh. 11 - Prob. 47PCh. 11 - Prob. 48PCh. 11 - Prob. 49PCh. 11 - Prob. 50PCh. 11 - Prob. 51PCh. 11 - Prob. 52PCh. 11 - Prob. 53PCh. 11 - Prob. 54PCh. 11 - Prob. 55PCh. 11 - Prob. 56PCh. 11 - Prob. 57PCh. 11 - Prob. 58PCh. 11 - Prob. 59PCh. 11 - Prob. 60PCh. 11 - Prob. 61PCh. 11 - Prob. 62PCh. 11 - Prob. 63PCh. 11 - Prob. 64PCh. 11 - Prob. 65PCh. 11 - Prob. 66PCh. 11 - Prob. 67GPCh. 11 - Prob. 68GPCh. 11 - Prob. 69GPCh. 11 - Prob. 70GPCh. 11 - Prob. 71GPCh. 11 - Prob. 72GPCh. 11 - Prob. 73GPCh. 11 - Prob. 74GPCh. 11 - Prob. 75GPCh. 11 - Prob. 76GPCh. 11 - Prob. 77GPCh. 11 - Prob. 78GPCh. 11 - Prob. 79GPCh. 11 - Prob. 80GPCh. 11 - Prob. 81GPCh. 11 - Prob. 82GPCh. 11 - Prob. 83GPCh. 11 - Prob. 84GPCh. 11 - Prob. 85GPCh. 11 - Prob. 86GPCh. 11 - Prob. 87GPCh. 11 - Prob. 88GP

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