Question

A particle executes simple harmonic motion of amplitude $A$. At what distance from the mean position is its kinetic energy equal to its potential energy?

• A. $0.81A$
• B. $0.71A$
• C. $0.41A$
• D. $0.91A$

Answer 1

The correct option is B $0.71A$

Distance form position can be represented as

$x=Asinwt....\left(i\right)$

By Differentiating wrt t, we get

$V=Awcoswt....\left(ii\right)$

we know that,

$KE=\frac{1}{2}m{v}^2$

$PE=\frac{1}{2}K{X}^2$ where $k=m^2\left(w^2\right)$

Then,

when $KE=PE$

$\frac{1}{2}m{v}^2=\frac{1}{2}K{X}^2\Rightarrow V^2=(WX{)}^2...(iii)$

From (i), (ii) and (iii)

$\left(AWcosw{t}^2\right)=(WAsinwt{)}^2$

$\Rightarrow (coswt{)}^2=(sinwt{)}^2\Rightarrow ta{n}^{2}wt=1$

So, $1\times 1=A/sinwt/=A\frac{1}{\sqrt{2}}=0.71A$