Question

The shortest distance traveled by a particle executing S.H.M from the mean position is $2s$ is equal to $\left( {\dfrac{{\sqrt 3 }}{2}} \right)$times its amplitude. Determine its time period.

Answer 1

Hint: The motion in which the body’s displacement is directly proportional to the restoring from its mean position is called simple harmonic motion. The restoring force’s direction is always towards the mean position. The simple harmonic motion equation will be useful to solve the given problem.
Formula used:
$ \Rightarrow x = A\sin \left( {\omega t} \right)$
Where $t$ is the time period and $\omega $ is the displacement.

Complete answer:
A particle is executing a simple harmonic motion from the mean position of $2s$ at the shortest distance traveled by the particle. The mean position is equal to the amplitude of $\left( {\dfrac{{\sqrt 3 }}{2}} \right)$times.
To determine the time period the simple harmonic motion equation will be helpful. The simple harmonic motion equation is,
$ \Rightarrow x = A\sin \left( {\omega t} \right)$
Where $t$ is the time period and $\omega $ is the displacement. The value of time is given as $2s$ and the value of $x$ is given as $\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ of the amplitude. Substitute the values.
$ \Rightarrow \dfrac{{\sqrt 3 }}{2}A = A\sin \left( {\omega 2} \right)$
The formula for the Time period is,
$ \Rightarrow T = \dfrac{{2\pi }}{\omega }$
Therefore, the displacement is given as,
$ \Rightarrow \omega = \dfrac{{2\pi }}{T}$
Substitute the value of omega in the simple harmonic equation.
$ \Rightarrow \dfrac{{\sqrt 3 }}{2}A = A\sin \left( {\dfrac{{2\pi }}{T}2} \right)$
Cancel out the common terms.
$ \Rightarrow \dfrac{{\sqrt 3 }}{2} = \sin \left( {\dfrac{{2\pi }}{T}2} \right)$
Simplify the given equation.
$ \Rightarrow \dfrac{{\sqrt 3 }}{2} = \sin \left( {\dfrac{{4\pi }}{T}} \right)$
We know that $\sin {60^ \circ }$is $\dfrac{{\sqrt 3 }}{2}$. $\sin {60^ \circ }$ can also be written as $\sin \dfrac{\pi }{3}$. Substitute in the equation.
$ \Rightarrow \sin \dfrac{\pi }{3} = \sin \left( {\dfrac{{4\pi }}{T}} \right)$
Cancel out the common terms and simplify the equation so we get the value of the time period.
$ \Rightarrow \dfrac{1}{3} = \left( {\dfrac{4}{T}} \right)$
$ \Rightarrow T = 4 \times 3$
$ \Rightarrow T = 12s$
Therefore, the time period of the particle that executes simple harmonic motion is $12s$.

Note:
The simple harmonic motion is very much useful in determining the characteristics of certain materials that execute this particular motion. It also forms an important tool in understanding the sound waves, alternating currents, and light wave characteristics. This motion at different frequencies of different oscillatory can be expressed as the superposition of several harmonic motions.