Question

A particle executes $SHM$ on a line $8cm$ long. Its $K.E$ and $P.E$ will be equal when its distance from the mean position is:

• A. $4cm$
• B. $2cm$
• C. $2\sqrt{2}cm$
• D. $\sqrt{2}cm$

Answer 1

The correct option is C $2\sqrt{2}cm$

SHM

$r=8cm$ because

$K.E=P.E$

So,

$\left(\frac{1}{2}\right)m{w}_2{y}_2{r}_2-y_2=y_2$

$\Rightarrow 2y^2=r^2\Rightarrow y$

$\to \frac{r}{v_2}=\frac{8}{v_2}=4\sqrt{2}cm$

From the mean position