Question

The periodic time of a body executing SHM is $4\mathrm{s}$. After how much interval from time $\mathrm{t}=0$, its displacement will be half of its amplitude?

Answer 1

Step 1: Given information.

Periodic time, $\mathrm{T}=4\sec$

Step 2: To find, After how many intervals from the time $\mathrm{t}=0$ its displacement will be half of its amplitude.

Step 3: Calculation.

The displacement equation in simple harmonic motion is given by,

$\mathrm{y}=\mathrm{asin\omega t}$…(equation $1$)

If $\mathrm{y}=\frac{\mathrm{a}}{2}$

By putting the value of y in the equation $1$, we get

$$\begin{gathered} \frac{\mathrm{a}}{2}=\mathrm{asin\omega t} \\ \mathrm{sin\omega t}=0.5 \\ \mathrm{\omega t}={\sin }^{-1}(0.5) \\ =\frac{\mathrm{\pi }}{6} \\ \frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}=\frac{\mathrm{\pi }}{6} \\ \mathrm{t}=\frac{\mathrm{T}}{12} \\ =\frac{4}{12} \\ =\frac{1}{3}\sec \end{gathered}$$

Therefore, the time interval is $\frac{1}{3}\sec$.