An aluminum sphere weighing 130 lbf, whereupon the spring is stretch 2.5 ft

Bacille John Purca

Bacille John Purca

Answered question

2022-05-18

 

 

Answer & Explanation

user_27qwe

user_27qwe

Skilled2022-07-02Added 375 answers

Initially, the spring is displaced by   = 2.5 ft

The weight of the ball, Mg = 130 lbf

 g = 32.17 ft/s 2

 M = 4.04 lb

From FBD,

 

Mg = K

 130 = K * 2.5

 K = 52 lbf / ft

Now, Natural frequency = ωn =  KM=  524.04 = 3.588 rad/s

Now, The equation of motion of the spring-mass system is,

x = A Sin ( ωnt ) , where x = displacement of the ball from the mean or equillibrium position in inch

A = maximum displacement from the mean position in inch

t = Time at which displacement is to be calculated in s

(ωnt ) is angular displacement in rad

Given,A = 6 inch

 x = 6 Sin ( ωnt )

x = 6 Sin ( 3.588 t ) ..........The equation for position of the ball from mean position at time t

b) At t = π12 s

The position of the ball is, x = 6 * Sin ( 3.588 * π12 )

 x = 4.84 inch

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