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.

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 .

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.