Question

Displacement-time equation of a particle executing SHM is $x = A\sin (\omega t + \dfrac{\pi }{6})$ time taken by the particle to go directly from $x = - \dfrac{A}{2}$ to $x = + \dfrac{A}{2}$ is:
(A) $\dfrac{\pi }{{3\omega }}$
(B) $\dfrac{\pi }{{2\omega }}$
(C) $\dfrac{{2\pi }}{\omega }$
(D) $\dfrac{\pi }{\omega }$

Answer 1

Hint
As we know that we have an equation of simple harmonic motion, and values of displacement are also given, then we try to simply put their values in the equation, and try to calculate two different values of time taken. After that difference in time taken will give us the time interval.

Complete step by step answer
Here $A$ is amplitude, $\omega$ is angular frequency
SHM equation, $x = A\sin (\omega t + \dfrac{\pi }{6})$
Points of displacement are $x = - \dfrac{A}{2}$ and $x = + \dfrac{A}{2}$
Let the time taken when $x = - \dfrac{A}{2}$ is ${T_1}$
And the time taken when $x = + \dfrac{A}{2}$ is ${T_2}$
Step by step solution:
We will put the value of $x = - \dfrac{A}{2}$ and $x = + \dfrac{A}{2}$ in a displacement-time equation.
$\Rightarrow x = A\sin (\omega t + \dfrac{\pi }{6})......(1)$
Put the value of $x = - \dfrac{A}{2}$ and $t = {T_1}$ in equation $(1)$
$\Rightarrow - \dfrac{A}{2} = A\sin (\omega {T_1} + \dfrac{\pi }{6})$
$A$ Will cancel out and multiply by ${\sin ^{ - 1}}$ on both side
$\Rightarrow - {\sin ^{ - 1}}(\dfrac{1}{2}) = \omega {T_1} + \dfrac{\pi }{6}$
Put value of ${\sin ^{ - 1}}(\dfrac{1}{2}) = \dfrac{\pi }{6}$
$\Rightarrow - \dfrac{\pi }{6} = \omega {T_1} + \dfrac{\pi }{6}$
Take only ${T_1}$ on left side, remaining on right side
$\Rightarrow {T_1} = - \dfrac{\pi }{{3\omega }}......(2)$
Here, we get value of ${T_1}$
Now, put the value of $x = + \dfrac{A}{2}$ and $t = {T_2}$ in equation $(1)$
$\Rightarrow \dfrac{A}{2} = A\sin (\omega {T_2} + \dfrac{\pi }{6})$
$A$ Will cancel out and multiply by ${\sin ^{ - 1}}$ on both side
$\Rightarrow {\sin ^{ - 1}}(\dfrac{1}{2}) = \omega {T_2} + \dfrac{\pi }{6}$
Put value of ${\sin ^{ - 1}}(\dfrac{1}{2}) = \dfrac{\pi }{6}$
$\Rightarrow \dfrac{\pi }{6} = \omega {T_2} + \dfrac{\pi }{6}$
$\Rightarrow {T_2} = 0......(3)$
As we know that time interval is equal to the difference of time taken.
So, the time interval by the particle:
$\Rightarrow T = {T_2} - {T_1}......(4)$
Put the value from equation $(2)$ and $(3)$ in equation $(4)$
$\Rightarrow T = 0 - ( - \dfrac{\pi }{{3\omega }}) = \dfrac{\pi }{{3\omega }}$
Hence the answer is option $(A)$ .

Additional Information
An oscillatory motion is called a simple harmonic motion, and it is directly proportional to the displacement of the object and it is opposite to the direction of the displacement of the object. Since the motion is periodic, it is defined by periodic functions such as sine, and cosine. The general equation of a simple harmonic motion is given below:
x(t)=Asin(ωt+ϕ)

Note
We should put values in simple harmonic motion equations carefully, and find out the time taken in both displacement cases, then the difference of both times will give us the time interval which we want.