A particle executes simple Harmonic Motion with amplitude of 4cm. At the mean position, the velocity of the particle is 10cm/sec. The distance of the particle from the mean position when its speed becomes 5cm/s is
A. $\sqrt 3 {\text{ cm}}$
B. $\sqrt 5 {\text{ cm}}$
C. $ \pm 2(\sqrt 3 {\text{) cm}}$
D. $2(\sqrt 5 {\text{) }}cm$

Hint: To solve this question, we should take use of the formula in which the velocity of the particle, amplitude of the particle and the distance of the particle from the mean position is related. Then from simple calculations, we will find the required distance.

Complete Step-by-Step solution:
The amplitude of the particle executing Simple harmonic Motion is given in the question
$ \Rightarrow A = 4cm$
Now, the relation between the amplitude of the particle executing SHM, the velocity of the particle and the distance of the particle from the mean position is given by the formula
$ \Rightarrow v = \omega \sqrt {{A^2} - {x^2}}$, where,
$v$ = velocity of the particle
$\omega$ = angular frequency of the particle
$A$ = Amplitude of the particle
$x$ = distance of the particle from the mean position

Now, according to the question, the velocity of the particle is 10 cm/s when it is at mean position
$\therefore x = 0{\text{ and v = 10}}$
Applying the formula, we get
\[
   \Rightarrow 10 = \omega \sqrt {{A^2} - {x^2}} \\
   \Rightarrow 10 = \omega \sqrt {{4^2} - {0^2}} \\
   \Rightarrow 10 = \omega (4) ……………….. 1 \\
\]
It is also given that the speed of the particle is 5cm at a distance x. It can be calculated as
$
   \Rightarrow 5 = \omega \sqrt {{A^2} - {x^2}} {\text{ ………………. 2}} \\
    \\
$
Dividing equation 1 with 2, we get
\[ \Rightarrow \dfrac{{10}}{5} = \dfrac{{\omega (4)}}{{\omega \sqrt {16 - {x^2}} }}\]
$
   \Rightarrow 2 = \dfrac{4}{{\sqrt {16 - {x^2}} }} \\
   \Rightarrow \sqrt {16 - {x^2}} = 2 \\
$
Squaring both sides
$
   \Rightarrow 16 - {x^2} = 4 \\
   \Rightarrow {x^2} = 12 \\
   \Rightarrow x = \pm 2\sqrt 3 {\text{ cm}} \\
$
$\therefore $ Option C is the correct option.

Note-The mean position in the Simple Harmonic Motions means the position where the particle comes to rest, that is, the net force on the particle is zero. In the above question, the distance where the speed of the particle becomes 5cm/s is $ \pm 2\sqrt 3 $ cm from the mean position. It means that there are two positions in the opposite direction at which the particle speed becomes 5cm/s.