Question

A particle executes simple harmonic motion with frequency 2.5 Hz and amplitude 2 m. The speed of the particle 0.3 s after crossing the equilibrium position is:

• A. Zero
• B. $2\pi$ m/s
• C. $4\pi$ m/s
• D. $\pi$ m/s

Answer 1

The correct option is A Zero

$T=\frac{1}{F}=\frac{1}{2.5}=0.4$ s
$t=0.3s=\frac{3T}{4}$
At the given time, particle win be in its extreme position if at t=0 it crosses the mean position.