Question

A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then stretched an additional 1 inch and released. Find the equation of motion, the amplitude, and the period. Neglect friction.

Answer 1

Step-by-step explanation:

angular frequency of resulting SHM ω = √(k/m

ω =√(100x32/1

ω = 56.57

Period T

ω = 2π /T = 56.57

T = 2π / 56.57

= .11 s .

spring is stretched by 1 inch so it will become the amplitude of frequency

A = 1 inch

when t = 0 , x ( displacement ) = A = 1 inch

equation of motion

x ( t ) in inch = A cosω t

= 1 cos 56.57t

x ( t ) in inch = 1 cos 56.57 t .