Question

Velocity at mean position of a particle executing SHM is v, then velocity of the particle at a distance equal to half of the amplitude is

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

Answer 1

Answer: (A)

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

At mean position $v=a\omega$
Velocity at half the amplitude is
$v^{\prime }=\omega \sqrt{a^2-y^2}$
$=\omega \sqrt{a^2-\frac{a^2}{4}}=\frac{\sqrt{3}}{2}a\omega =\frac{\sqrt{3}}{2}v$