Question

A simple harmonic motion has an amplitude $A$ and time period $T$ . Find the time required by it to travel directly from $x = 0$ to $x = \dfrac{A}{2}$ .

Answer 1

Hint: Use the formula of the displacement of the simple harmonic motion and substitute the displacement in it. From the simplified relation substitute the angular frequency formula to find the time required to travel the displacement of half the amplitude.
Useful formula:
(1) The formula of the displacement of the simple harmonic motion is given by
$x = A\sin \omega t$
Where $x$ is the displacement of the wave, $A$ is the amplitude of the simple harmonic motion and $\omega $ is the angular frequency of the wave and $t$ is the time taken for the displacement.
(2) The formula of the angular frequency of the simple harmonic motion is given by
$\omega = \dfrac{{2\pi }}{T}$
Where $\omega $ is the angular frequency of the wave and $T$ is the time period of the wave.

Complete step by step solution:
It is given that the amplitude and the time period of the simple harmonic motion is $A$ and $T$ respectively.
Initial position is $x = 0$ and the final position is $x = \dfrac{A}{2}$
Using the formula of the displacement of the simple harmonic motion,
$x = A\sin \omega t$
Substituting the displacement in the above equation, we get
$
  \dfrac{A}{2} = A\sin \omega t \\
  \omega t = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\
  \omega t = \dfrac{\pi }{6} \\
 $
Substituting the formula of the angular frequency in the above step, we get
$\dfrac{{2\pi }}{T}t = \dfrac{\pi }{6}$
By cancelling the similar terms in the above equation,
$t = \dfrac{T}{{12}}$

Hence the time required to travel from one position to the other is $\dfrac{T}{{12}}$ .

Note: In the simple harmonic motion, the restoring force is equal to the object magnitude and it is the periodic motion. The sine wave is an example of this type of motion. Time period is the reciprocal of the angular frequency. Remember the formula of the displacement of simple harmonic motion.