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Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change value from half the amplitude to the amplitude.

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#### Solution

As per the conditions given in the question,

\[y_1 = \frac{A}{2}; \]

\[ y_2 = A\]

(for the given two positions)

Let *y*_{1}_{ }and *y*_{2} be the displacements at the two positions and *A *be the amplitude.

Equation of motion for the displacement at the first position is given by,

*y*_{1} = *A*sinω*t*_{1}

As displacement is equal to the half of the amplitude,

\[\frac{A}{2} = A \sin \omega t_1\]

\[\Rightarrow \sin \omega t_1 = \frac{1}{2}\]

\[ \Rightarrow \frac{2\pi \times t_1}{T} = \frac{\pi}{6}\]

\[ \Rightarrow t_1 = \frac{T}{12}\]

The displacement at second position is given by,*y*_{2} = *A* sin *ωt*_{2}

As displacement is equal to the amplitude at this position,

⇒ *A* = *A* sin *ωt*_{2}

⇒ sinω*t*_{2} = 1

\[\Rightarrow \omega t_2 = \frac{\pi}{2}\]

\[ \Rightarrow \left( \frac{2\pi}{T} \right) t_2 = \frac{\pi}{2} \left( \because \sin \frac{\pi}{2} = 1 \right)\]

\[ \Rightarrow t_2 = \frac{T}{4}\]

\[ \therefore t_2 - t_1 = \frac{T}{4} - \frac{T}{12} = \frac{T}{6}\]

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