Question

A particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude.

• A. $\frac{T}{2}$
• B. $\frac{T}{4}$
• C. $\frac{T}{8}$
• D. $\frac{T}{12}$

Answer 1

The correct option is D

$\frac{T}{12}$

Step 1: Given data

A particle executes a simple harmonic motion of time period $T$.

Step 2: To find

Time taken by the particle to go directly from its mean position to half the amplitude.

Step 3: Definition and calculation

Simple harmonic motion (SHM)

1. Simple harmonic motion is defined as a motion in which the restoring force is directly proportional to the displacement of the body from its mean position.
2. Its direction is always towards the mean position.

Displacement of an SHM is given by,

$x=A\sin \omega t$, where $A$ is the amplitude of oscillation, $t$ is time and $\omega$ is the angular frequency.

At mean position,

$t=0andx=0$

At half amplitude,

$$\begin{gathered} \frac{A}{2}=A\sin \omega t \\ \Rightarrow \frac{1}{2}=\sin \omega t \\ \Rightarrow \omega t={\sin }^{-1}\left(\frac{1}{2}\right) \\ \Rightarrow \omega t=\frac{\mathrm{\pi }}{6} \\ \Rightarrow \frac{2\mathrm{\pi }}{T}\times t=\frac{\mathrm{\pi }}{6}\left(\because \omega =\frac{2\mathrm{\pi }}{T}\right) \\ \Rightarrow t=\frac{T}{12} \end{gathered}$$

Therefore, the time taken for the particle to go from mean position to half amplitude is $\frac{T}{12}$.

Hence, option D is correct.