Question 318: Oscillations

Question

The amplitude of a particle executing S.H.M. is $$3 \mathrm{~cm}$$. The displacement at which its kinetic energy will be $$25 \%$$ more than the potential energy is

Accepted Answer

$$\begin{aligned}
& 	ext { Given }: \mathrm{K} \cdot \mathrm{E}=\mathrm{P} \cdot \mathrm{E}+\frac{25}{100} \cdot \mathrm{P} \cdot \mathrm{E} \
& \mathrm{K} \cdot \mathrm{E}=\mathrm{P} \cdot \mathrm{E}+\frac{1}{4} \mathrm{P} \cdot \mathrm{E} \
& 	ext { We know K.E }=\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2ight) 	ext { and P.E }=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2 \
& 	herefore \quad \mathrm{K} \cdot \mathrm{E}=\frac{5}{4} \mathrm{P} \cdot \mathrm{E} \
& \frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2ight)=\frac{5}{4}\left(\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2ight) \
& \mathrm{A}^2-\mathrm{x}^2=\frac{5}{4} \mathrm{x}^2 \
& \mathrm{~A}^2=\frac{9}{4} \mathrm{x}^2 \
	herefore \quad & \mathrm{~A}=\frac{3}{2} \mathrm{x} \
	herefore \quad & \mathrm{x}=\mathrm{A} 	imes \frac{2}{3}=2 \mathrm{~cm}
\end{aligned}$$

Answer

$$1 \mathrm{~cm}$$

Answer

$$2 \mathrm{~cm}$$

Answer

$$3 \mathrm{~cm}$$

Answer

$$4 \mathrm{~cm}$$

Oscillations

The amplitude of a particle executing S.H.M. is $3 \mathrm{~cm}. The displacement at which its kinetic energy will be 25 \%$ more than the potential energy is

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