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Kamila [148]
3 years ago
13

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?
Physics
1 answer:
____ [38]3 years ago
3 0

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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