Question 301: Oscillations

Question

The time taken by a particle executing simple harmonic motion of period '$$\mathrm{T}$$', to move from the mean position to half the maximum displacement is

Accepted Answer

$$\begin{aligned}
& 	ext { } \mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}=\mathrm{A} \sin \left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}ight) 	ext { where } \mathrm{x}=\mathrm{A} / 2 \
& 	herefore \frac{\mathrm{A}}{2}=\mathrm{A} \sin \left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}ight) \
& 	herefore \frac{1}{2}=\sin \frac{\pi}{6}=\sin \left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}ight) \
& 	herefore \frac{2 \pi \mathrm{t}}{\mathrm{T}}=\frac{\pi}{6} \
& 	herefore \mathrm{t}=\frac{\mathrm{T}}{12} \mathrm{~s}
\end{aligned}$$

Answer

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

Answer

$$\frac{\mathrm{T}}{2} \mathrm{~s}$$

Answer

$$\frac{\mathrm{T}}{4} \mathrm{~s}$$

Answer

$$\frac{\mathrm{T}}{6} \mathrm{~s}$$

Oscillations

The time taken by a particle executing simple harmonic motion of period '$\mathrm{T}$', to move from the mean position to half the maximum displacement is

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