Question

A 0.20-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?

Answer 1

Answer:

$0.57 ms^{-1}$

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 $w = \sqrt{\frac{k}{m} } \\w = \sqrt{\frac{40}{0.20} }\\w = 14.14 rads^{-1}$

$v$ = Speed of the block as it pass the equilibrium point

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

$v = A w\\v = (0.04) (14.14)\\v = 0.57 ms^{-1}$