Question

A particle execute S.H.M from the mean position. its amplitude is $A$, its time period is "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. $\frac{3A}{\sqrt{A}}$

Answer 1

Answer: (A)

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

For a SHM velocity, $v=\omega \sqrt{A^2-x^2}$. . . . . . .(1)

The maximum velocity is at $x=0$

$v_m=\omega A$

Now, $v=\frac{\omega A}{2}$

From equation (1),

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

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

$x^2=\frac{3A^2}{4}$

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

The correct option is A.