Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a spring whose

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Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a spring whose spring constant is 61.6 N/m. An observer is traveling at a speed of 2.79 × 108 m/s relative to the fixed end of the spring. What does this observer measure for the period of oscillation?

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Thông Đạt 5 years 2021-07-14T04:19:58+00:00 1 Answers 17 views 0

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  1. minhkhoi
    0
    2021-07-14T04:21:49+00:00
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    Explanation:

    Given that,

    Mass of the object, m = 7.11 kg

    Spring constant of the spring, k = 61.6 N/m

    Speed of the observer, v=2.79\times 10^8\ m/s

    We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

    t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s

    Time period of oscillation measured by the observer is :

    t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s

    So, the time period of oscillation measured by the observer is 5.79 seconds.

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