Question

A particle is performing SHM with an amplitude $A$ and angular frequency $\omega$. Find the displacement of the particle from the mean position where the speed of the particle becomes half of the maximum speed.

• A. $x=\pm \frac{A}{2}$
• B. $x=\frac{A}{\sqrt{2}}$
• C. $x=\pm \frac{\sqrt{3}A}{2}$
• D. $x=\frac{-A}{\sqrt{2}}$

Answer 1

The correct option is C $x=\pm \frac{\sqrt{3}A}{2}$

Let the displacement of the particle from mean position be $x$.
Given that, $v=\frac{v_{max}}{2}$
But we know that for a particle executing SHM, the speed of the particle is given by
$v=\pm \omega \sqrt{A^2-x^2}.......\left(1\right)$
And also that, $v_{max}=A\omega$ [at mean position]
From $\left(1\right)$ we can say that,
$\frac{A\omega }{2}=\pm \omega \sqrt{A^2-x^2}\Rightarrow \frac{A}{2}=\pm \sqrt{A^2-x^2}$
Squaring on both sides, we get
$A^2-x^2=\frac{A^2}{4}\Rightarrow x^2=\frac{3A^2}{4}\Rightarrow x=\pm \frac{\sqrt{3}A}{2}$
Thus, option (c) is the correct answer.