1) A steel ball weighing 128 pounds is suspended from a spring, whereupon the sp

Question

1) A steel ball weighing 128 pounds is suspended from a spring, whereupon the spring is stretched 2 feet from its natural length. The ball is started in motion with no initial velocity by displacing it 6 inches above the equilibrium position. For this problem, the frequency is [ ] and the period is [ ].

2) A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. The mass is set into motion in a medium that imparts a damping force numerically equal to 16 times the velocity. If the mass is pulled down two centimeters and released, the solution for the position is.....

Explanation / Answer

k = 64 lb/ft for spring

position above equilirium = x
at t = 0 x = .5 ft
so
x = .5 cos 2 pi t/T
we need to find T, the period
f = -kx = m a
but a = .5 (2 pi/T)^2(-cos2 pi/T) =-(2pi/T)^2 x
so
-k x = -(2 pi/T)^2 x
2 pi/T = sqrt(k/m)
m = 128/32 = 4
so
2 pi/T = sqrt (64/4) = 4
therefore
x = .5 cos 4 t

at t = pi/12
x = .5 cos pi/3 = .5 cos 60deg = .25

w = 2 pi f = 2 pi/T = 4 radians/sec
f = w/2pi = 2/pi
T = pi/2

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