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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.

Respuesta :

Answer:

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 .

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