For what value of displacement

Question:

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

  1. $x=0$

  2. $x=\pm A$

  3. $x=\pm \frac{A}{\sqrt{2}}$

  4. $x=\frac{A}{2}$


Correct Option: , 3

Solution:

$\mathrm{KE}=\mathrm{PE}$

$\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\frac{1}{2} m \omega^{2} x^{2}$

$A^{2}-x^{2}=x^{2}$

$2 x^{2}=A^{2}$

$x=\pm \frac{A}{\sqrt{2}}$

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