Question

the PE Of SHM is 0.1sec after crossing the mean position is 1/4of it's total energy. then the period of it's? ​

Answer 1

Answer:

T=0.6s

Explanation:

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Total Energy of SHM is given by $\frac{1}{2}$mω²A²

and PE is given by  $\frac{1}{2}$mω²(A²-x²)

Now;

   $\frac{1}{2}$mω²(A²-x²)= $\frac{1}{4}$($\frac{1}{2}$mω²A²)

=>A²-x²=$\frac{A^{2} }{4}$

=>x=$\frac{\sqrt[]{3}A }{2}$  -----(1)

now ;   x = Asin(ωt)

         $\frac{\sqrt[]{3}A }{2}$=Asin(ωt)

         ∴ωt=$\frac{\pi }{3}$

            $\frac{2\pi }{T}$(0.1)=$\frac{\pi }{3}$

           ∴T=3/5=0.6s

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