The displacement of a simple harmonic motion of amplitude 6 cm when its kinetic energy is equal to its…
The displacement of a simple harmonic motion of amplitude 6 cm when its kinetic energy is equal to its potential energy is
$2 \sqrt{2} \mathrm{~cm}$
2 cm
$3 \sqrt{2} \mathrm{~cm}$
$\frac{3}{\sqrt{2}} \mathrm{~cm}$
Solution
Given, amplitude of oscillation, a = 6 cm
Let displacement be x cm.
When kinetic energy is equal to potential energy,
then
$\begin{array}{rlrl} & \frac{1}{2} m \omega^2\left(a^2-x^2\right) & =\frac{1}{2} m \omega^2 x^2 \\ \Rightarrow & a^2-x^2 & =x^2 \Rightarrow 2 x^2=a^2 \\ \Rightarrow & 2 x^2 & =36 \Rightarrow x^2=18 \\ \Rightarrow & & x & =3 \sqrt{2} \mathrm{~cm}\end{array}$