A0.20-kg block rests on a frictionless level surface and is attached to a horizontally aligned spring with a spring constant of 40 n/m. the block is initially displaced 4.0 cm from the equilibrium point and then released to set up a simple harmonic motion. what is the speed of the block when it passes through the equilibrium point?

Explanation:

k = spring constant of the spring = 40 Nm⁻¹

A = amplitude of the simple harmonic motion = 4 cm = 0.04 m

m = mass of the block attached to spring = 0.20 kg

w = angular frequency of the simple harmonic motion

Angular frequency of the simple harmonic motion is given as

= Speed of the block as it pass the equilibrium point

Speed of the block as it pass the equilibrium point is given as

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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