Answer to Question #200434 in Calculus for Raghav

Question #200434

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

seconds. Find the maximum velocity of the particle.

Expert's answer

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

\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.

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