Question

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.

Answer 1

As per the conditions given in the question,

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

\[ y_2 = A\]

    (for the given two positions)

Let y₁ and y₂ 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,
 y₁ = Asinωt₁
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,
y₂ = A sin ωt₂
As displacement is equal to the amplitude at this position,
⇒        A = A sin ωt₂
⇒ sinωt₂ = 1

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

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

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

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