Question

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

• A. $x=\frac{A}{2}$
• B. $x=0$
• C. $x=\pm A$
• D. $x=\pm \frac{A}{\sqrt{2}}$

Answer 1

The correct option is D

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

Step 1. Given Data:

The kinetic energy and potential energy of a simple harmonic oscillation become equal

That is, $K.E=P.E$

Step 2. Find the value of displacement:

The kinetic energy of SHM is given by,

$K.E=\frac{1}{2}k\left(A^2-x^2\right)$

Where, $k=$spring constant

$A=$amplitude

$x=$displacement

The potential energy of SHM is,

$P.E=\frac{1}{2}k\left(x^2\right)$

Now, according to the question,

$K.E=P.E$

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

$\Rightarrow$ $\left(A^2-x^2\right)=\left(x^2\right)$

$\Rightarrow$ $A^2=\left(2x^2\right)$

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

Hence, option D is correct