Question 1142357
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The function is    F(t) = {{{6*cos((2pi/3)*t)+10}}}.


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The given function describes periodic oscillations of the mass around its averaged position at 10 inches from the ceiling 
with the amplitude of 6 inches and the period of 3 seconds.



(a)  At the moment t= 0, when the mass was released, its distance from the ceiling was  6 + 10 = 16 inches.    <U>ANSWER</U>



(b)  At its closest position from the ceiling, the distance of the mass from the ceiling is  10 - 6 = 4 inches.    <U>ANSWER</U>



(c)  First time the weight will be at 13 inches from the ceiling at the time moment t, when


     {{{6*cos((2pi/3)*t)+10}}} = 13,   or


     {{{6*cos((2pi/3)*t)}}} = 13 - 10 = 3,   or


     {{{cos((2pi/3)*t)}}} = {{{3/6}}} = {{{1/2}}}.


It means that at this moment


     {{{(2pi/3)*t}}} = {{{pi/3}}};


hence,


    t = {{{1/2}}} of a second.      <U>ANSWER</U>



(d)  The other time moments  t <= 6 seconds, when the mass will be at the distance of 13 inches from the ceiling

      are the OTHER solutions of the equation 


     {{{6*cos((2pi/3)*t)+10}}} = 13,   or


     {{{6*cos((2pi/3)*t)}}} = 13 - 10 = 3,   or


     {{{cos((2pi/3)*t)}}} = {{{3/6}}} = {{{1/2}}}.


     So, they are 


        t = 3 - {{{1/2}}} = 2.5 seconds;

        t = 3 + {{{1/2}}} = 3.5 seconds,  and

        t = 6 - {{{1/2}}} = 5.5 seconds.
</pre>


All questions are answered -- the problem is solved.