Question 267: Oscillations

Question

Starting from mean position, a body oscillates simple harmonically with a period ' $T$ '. After what time will its kinetic energy be $75 \%$ of the total energy? $\left(\sin 30^{\circ}=0.5\right)$

Accepted Answer

$$\begin{array}{ll} 
& \mathrm{KE}=75 \% 	ext { of TE } \
	herefore & \frac{1}{2} \mathrm{ma}^2 \omega^2 \cos ^2 \omega \mathrm{t}=\frac{75}{100} 	imes \frac{1}{2} \mathrm{ma}^2 \omega^2 \
	herefore & \cos ^2 \omega \mathrm{t}=\frac{3}{4} \
	herefore & \cos \omega \mathrm{t}=\frac{\sqrt{3}}{2} \
	herefore \quad & \omega \mathrm{t}=\frac{\pi}{6} \
	herefore & \frac{2 \pi}{\mathrm{~T}} \mathrm{t}=\frac{\pi}{6} \quad 	herefore \quad \mathrm{t}=\frac{\mathrm{T}}{12}
\end{array}$$

Answer

$\frac{\mathrm{T}}{8}$

Answer

$\frac{\mathrm{T}}{12}$

Answer

$\frac{\mathrm{T}}{16}$

Answer

$\frac{\mathrm{T}}{24}$

Oscillations

Starting from mean position, a body oscillates simple harmonically with a period ' T '. After what time will its kinetic energy be 75 \% of the total energy? \left(\sin 30^{\circ}=0.5\right)

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