Question

A 0.250 kg block attached to a light spring oscillates on a frictionless, horizontal table. The oscillation amplitude is A = 0.125 m and the block moves at 3.00 m/s as it passes through equilibrium at x = 0. (a) Find the spring constant, k.

Answer 1

The spring constant, k will be 18 N/m.The ratio of force to one unit of displaced length is known as the spring constant.

What is the spring constant?

Spring constant is defined as the ratio of force per unit displaced length.

Given data;

Mass of  block,m=0.250 k

Amplitude,A = 0.125 m

Velocity,V=3.00 m/

Spring constant,K=?

The formula for the spring constant when the spring is oscillate at the given amplitude A:

$\rm K = \frac{V_{max}^2 m}{A} \\\\ K = \frac{3.0 \times 0.250}{0.125} \\\\ K=18 \ N/m$

Hence, the spring constant, k will be 18 N/m

To learn more about the spring constant, refer:

brainly.com/question/4291098

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