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A particle executes simple harmonic motion with an amplitude of 2.18 cm.
At what positive displacement from the midpoint of its motion does its speed equal one half of its maximum speed?
Answer in units of cm.

Respuesta :

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is  :

                          v (max) = Aω

Speed v at any displacement y is given by  

[tex]v^{2}[/tex] = [tex]w^{2}[/tex] ([tex]A^{2}[/tex] - [tex]y^{2}[/tex])   ........................................................  i

And,

               v = [tex]\frac{1}{2}[/tex] v (max)  

          or,  2 × v = Aω     ....................................................   ii

Eliminating  ω from equations i and ii,

                       [tex]\frac{1}{4}[/tex] [tex]A^{2}[/tex]  [tex]w^{2}[/tex]  =  [tex]w^{2}[/tex]  ( [tex]A^{2}[/tex]  - [tex]y^{2}[/tex])

                     or, [tex]y^{2}[/tex] =  ([tex]\frac{3}{4}[/tex]) [tex]A^{2}[/tex]  =([tex]\frac{3}{4}[/tex]) [tex]2.18^{2}[/tex]

                    or,  y =  3.56 cm.

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