Question 1204497
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A mass is suspended by a spring such that it hangs at rest 0.5 m above the ground. 
The mass is raised 40 cm and released at time t= 0 s, causing it to oscillate sinusoidally. 
If the mass returns to the high position every 1.2 s, determine the height 
of the mass above the ground at t= 0.7 s.
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In this problem, the height h = 0.5 m above the ground is the equilibrium position.

Under given conditions, the mass will oscillate harmonically around this position 
with the amplitude of 0.4 m.


At given condition, the mass position is described by this equation

    h(t) = {{{0.5 + 0.4*cos((2pi/1.2)*t)}}},   t >= 0,


where h(t) is the height of the mass over the ground, in meters, t is the time, in seconds.


At t = 0.7 s, the height over the ground is  h(0.7) = {{{0.5 + 0.4*cos((6.28/1.2)*0.7)}}} = 0.15322 of a meter.
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Solved.