Question

A particle execute $SHM$ from the mean position$.$ its amplitude is $A$ its time period $T.$ At what displacement its speed is half of its maximum speed$?$

• A. $\frac{\sqrt{3}A}{2}$
• B. $\frac{\sqrt{2}}{3}A$
• C. $\frac{2A}{\sqrt{3}}$
• D. $\frac{3A}{\sqrt{2}}$

Answer 1

Velocity of simple harmonic motion$:$

$v=\omega \sqrt{A^2-y^2}$

$V_{max}=\omega A$

Given that$:$

$V=\frac{V_{max}}{2}$

$\omega \sqrt{A^2-y^2}=\frac{\omega A}{2}$

$\sqrt{A^2-y^2}=\frac{A}{2}$

$A^2-y^2=\frac{A^2}{4}$

$y^2=A^2-\frac{A^2}{4}=\frac{3A^2}{4}$

$y=\frac{\sqrt{3}}{2}A$

Hence,

option $\left(A\right)$ is correct answer.