Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of the block as it passes through the equilibrium position is 62 cm/s. What is the angular frequency of the block's motion

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

answer: i pick a

explanation:

kepler used the data from tycho blahe (his mentor) to construct the laws of planetary motion.