Question 1142780
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<pre>
Consider a harmonic oscillation


    x = a*sin(bt)      (1)


with the unknown amplitude " a " and unknown coefficient " b ".


For such oscillations,  the velocity is  


    v = x'(t) = ab*cos(bt)


and the acceleration is 


    a = x''(t) = -ab^2*sin(bt).


Also notice that for this harmonic motion the initial position is  x = 0  at t= 0.


In your case, take  ab = 10  and ab^2 = -25.


Then you will have <U>EXACTLY the described motion</U>.


From (2),  you have first  


    b = {{{(ab^2)/(ab)}}} = {{{(-25)/10}}} = -2.5,

and then

    a = {{{10/b}}} = {{{10/(-2.5)}}} = -4.


So, your motion is


    x = -4*sin(-2.5*t) = 4*sin(2.5t).

with

    v = x'(t) = 4*2.5*cos(2.5t) = {{{10*cos(2.5*(pi/15))}}} = {{{10*cos(pi/6))}}} = {{{10*(sqrt(3)/2))}}} = 8.66.


<U>ANSWER</U>.  The velocity after  {{{pi/15}}} seconds is 8.66 m/s.
</pre>

Solved.


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To solve this problem successfully, you need to know at least one of two facts:


<pre>
    a)  Either to know from Math that the solution/solutions to the equation


            x''(t) = -ax


         are the functions  a*sin(bt),  a*cos(bt),    or



    b)  to know from Physics, that the motion where the returning force is proportional to the displacement from the equilibrium 

        taken with the opposite sign, is a harmonic motion.
</pre>


I know both these facts, and it facilitates the solution significantly.



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<U>Comment from student</U> : &nbsp;&nbsp;The answer given is &nbsp;5 m/s ?? 
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<U>My response</U> :  &nbsp;&nbsp;The answer is &nbsp;8.66 m/s