Answer to Question #200434 in Calculus for Raghav

Question

A particle is executing Simple Harmonic Motion of amplitude 6 m and period 3× 5

seconds. Find the maximum velocity of the particle.

Answer 1

$A=6\ m, T=3.5\ s$

$$\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{3.5}rad/s$$ $$v_{max}=A\omega=6(\dfrac{2\pi}{3.5}) \ m/s=\dfrac{24\pi}{7}\ m/s$$

The maximum velocity of the particle is $\dfrac{24\pi}{7}$ m/s.