A particle performing simple harmonic motion in such that it’s amplitude is 4 m and speed of particle at mean position is 10 m/s. Find the distance of particle from mean position where velocity became 5 m/s.

The correct option is 2√3 m.

To find the distance of the particle from the mean position where its velocity is 5 m/s, we can use the standard equations of motion for Simple Harmonic Motion (SHM).

1. Identify the given parameters:
Amplitude of the motion, $A=4\text{m}$
Maximum speed (speed at the mean position), $v_{\text{max}}=10\text{m/s}$
Velocity at the desired position, $v=5\text{m/s}$

2. Find the angular frequency ($\omega$):
The maximum velocity of a particle performing SHM is related to the angular frequency and amplitude by the formula:
$v_{\text{max}}=\omega A$
Substituting the given values:
$10=\omega ·4$
$\omega =\frac{10}{4}=2.5{\text{rad/s}}^{}$

3. Calculate the displacement ($x$) from the mean position:
The velocity of a particle in SHM at any displacement $x$ from the mean position is given by:
$v=\omega \sqrt{A^2-x^2}$
Substituting $v=5\text{m/s}$, $\omega =2.5{\text{rad/s}}^{}$, and $A=4\text{m}$:
$5=2.5\sqrt{4^2-x^2}$
Dividing both sides by 2.5:
$2=\sqrt{16-x^2}$
Squaring both sides:
$2^2=16-x^2$
$4=16-x^2$
Rearranging the equation to solve for $x^2$:
$x^2=16-4$
$x^2=12$
Taking the square root on both sides:
$x=\sqrt{12}=2\sqrt{3}\text{m}$

Thus, the distance of the particle from the mean position where its velocity becomes 5 m/s is 2√3 m.