Question

A particle executes simple harmonic motion with an amplitude of 2.18 cm.

Answer 1

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  

$v^{2}$ = $w^{2}$ ($A^{2}$ - $y^{2}$)   ........................................................  i

And,

               v = $\frac{1}{2}$ v (max)  

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

Eliminating  ω from equations i and ii,

                       $\frac{1}{4}$ $A^{2}$  $w^{2}$  =  $w^{2}$  ( $A^{2}$  - $y^{2}$)

                     or, $y^{2}$ =  ($\frac{3}{4}$) $A^{2}$  =($\frac{3}{4}$) $2.18^{2}$

                    or,  y =  3.56 cm.