Question

A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from equilibrium position is
\[\begin{align}
  & \text{A}\text{. }\dfrac{T}{4} \\
 & \text{B}\text{. }\dfrac{T}{8} \\
 & \text{C}\text{. }\dfrac{T}{12} \\
 & \text{D}\text{. }\dfrac{T}{2} \\
\end{align}\]

Answer 1

Hint: First understand about simple harmonic motion. Then using the mathematical equation of simple harmonic motion you can derive the time taken. Then match your solution with the correct option.

Complete step by step answer:
Simple harmonic motion abbreviated as SHM is the oscillating motion of anybody. It is a type of periodic motion that is, the motion which repeats itself after some time interval.
Characteristics of simple harmonic motion can be:
-Motion of an object considered is back and forth.
-Motion of a body is periodic in nature.
-Body is accelerated in direction pointing towards equilibrium and should be directly proportional to the displacement.
Equation of SHM is given as:
\[x=a\sin \omega t\]
where,
a is amplitude, $\omega $ is the angular frequency and t is time.
Now, according to question,
$x=\dfrac{a}{2}$
Thus, the above equation can be written as:
$\begin{align}
  & \dfrac{a}{2}=a\sin \omega t \\
 & \Rightarrow \sin \omega t=\dfrac{1}{2} \\
\end{align}$
But we know, $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$, the above equation we get,
$\omega t=\dfrac{\pi }{6}$
Here, $\omega =\dfrac{2\pi }{T}$.
Thus, we can write equation as:
$\begin{align}
  & \dfrac{\pi }{6}=\dfrac{2\pi t}{T} \\
 & \Rightarrow t=\dfrac{T}{12} \\
\end{align}$
So, option C is the correct answer.

Additional Information:
Few examples of simple harmonic motions can be:
-Motion of a pendulum
-Bungee jumping
-Motion of spring

Note: While calculating time period consider the correct value of cosine. Also, do not forget to take distance equal to half of amplitude as given in the equation. Any of the above mistakes will lead to incorrect answers.