A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially rel

Question

A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 4 inches below the equilibrium position.
(a) Find the position x of the mass at the times t = pi/12, pi/8, pi/6, pi/4, and 9pi/32 s. Use g = 32 ft/s^2 for the acceleration due to gravity. (b) what is the velocity of the mass when t = 3pi/16 s?
A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 4 inches below the equilibrium position.
(a) Find the position x of the mass at the times t = pi/12, pi/8, pi/6, pi/4, and 9pi/32 s. Use g = 32 ft/s^2 for the acceleration due to gravity. (b) what is the velocity of the mass when t = 3pi/16 s?

(a) Find the position x of the mass at the times t = pi/12, pi/8, pi/6, pi/4, and 9pi/32 s. Use g = 32 ft/s^2 for the acceleration due to gravity. (b) what is the velocity of the mass when t = 3pi/16 s?

Explanation / Answer

M = 20 pounds =9.072 kg

x = 6 inches = 0.1524 m

mg=kx

9.072*9.8 = k*0.1524

k=583.37

T = 2*pi(m/k)^1/2 =0.783 s

? = 2*pi/T = 8.024

y = Asin(?t +?)

A = 4 inches = 0.1016 m

Initial condition

t=0; y=-0.1016 = A

?t +? =-pi/2 , ? =-pi/2

t=T/4 y=0

?t = ?*2*pi/?*4 = pi/2

?t +? = pi/2 - pi/2 =0

a) y= 0.1016sin(8.024*t -pi/2)

All distances relative to equilibrium

t=pi/12

y = 0.054 m= 2.13 in

t=pi/8

y = 0.102 m = 4 in

t =pi/6

y = 0.049 m = 1.93 in

t = pi/4

y = -0.1016 m = - 4 in

t = 9pi/32

y = -0.070 m =2.76 in

b) v=8.024*0.1016cos(8.024*t -pi/2)

t =3*pi/16

v = -0.82 m/s = -2.69 ft/s

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