Question 321: Oscillations

Question

A particle is vibrating in S.H.M. with an amplitude of $$4 \mathrm{~cm}$$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

Accepted Answer

$$\begin{aligned}
	ext { K.E } & =\frac{1}{2} m \omega^2\left(A^2-x^2ight) 	ext { and } \
	ext { P.E } & =\frac{1}{2} m \omega^2 x^2 \
	herefore \quad 	ext { T.E } & =	ext { K.E }+ 	ext { P.E } \
& =\frac{1}{2} m \omega^2\left(A^2-x^2ight)+\frac{1}{2} m \omega^2 x^2 \
& =\frac{1}{2} m \omega^2 A^2 \
& 	ext { Given: K.E }=	ext { P.E } \
	herefore \quad A^2 & -x^2=x^2 \
	herefore \quad & A^2=2 x^2 \
	herefore \quad x & =\sqrt{\frac{A^2}{2}}=\frac{A}{\sqrt{2}} \
	herefore \quad x & =\frac{4}{\sqrt{2}}=2 \sqrt{2} ~\mathrm{c m}
\end{aligned}$$

Answer

$$1 \mathrm{~cm}$$

Answer

$$\sqrt{2} \mathrm{~cm}$$

Answer

$$2 \mathrm{~cm}$$

Answer

$$2 \sqrt{2} \mathrm{~cm}$$

Oscillations

A particle is vibrating in S.H.M. with an amplitude of $4 \mathrm{~cm}$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

Choose the correct answer: