PROBLEM 14.2: "If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. The object is displaced a distance 0.120 m from its equilibrium position and released with zero initial speed. Then after a time 0.800 s, its displacement is found to be a distance 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. a) Find the amplitude. b) Find the period. c) Find the frequency."

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xero099 xero099 · Answer 1 · 2020-03-31T03:27:50+02:00

Answer:

a) [tex]A = 0.120\,m[/tex], b) [tex]T = 1.6\,s[/tex], c) [tex]f = 0.625\,hz[/tex]

Explanation:

The known conditions of the motion are:

[tex]x(0) = 0.120\,m[/tex]

[tex]x(0.8) = - 0.120\,m[/tex]

a) The amplitude is the half of the difference between both positions:

[tex]A = \frac{x(0)-x(0.8)}{2}[/tex]

[tex]A = \frac{0.120\,m-(-0.120\,m)}{2}[/tex]

[tex]A = 0.120\,m[/tex]

b) The period of oscillation is equal to the double of the time between both positions:

[tex]T = 2\cdot \Delta t[/tex]

[tex]T = 2\cdot (0.8\,s)[/tex]

[tex]T = 1.6\,s[/tex]

c) The frequency of the oscillation is:

[tex]f = \frac{1}{T}[/tex]

[tex]f = \frac{1}{1.6\,s}[/tex]

[tex]f = 0.625\,hz[/tex]