A vertical spring with spring stiffness constant 305 N/m oscillates with an ampl
ID: 1435382 • Letter: A
Question
A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0 cm when 0.235 kghangs from it. The mass passes through the equilibrium point (y=0) with positive velocity at t=0. Positive direction of y-axis is downward.
What equation describes this motion as a function of time? (answer in terms of t, measured in seconds).
At what time will the spring stretch to its maximum length at first time?
At what time will the spring shrink to its minimum length at first time?
Explanation / Answer
since it starts at equilibrium and not full displacement, then the equation is as follows
y(t) = A * sin(w*t)
A = 28.0 cm = 0.28 m
w = sqrt(k/m) = sqrt(305 N/m / 0.235 kg) = 36.0 rad/s
It stretches to the max when y(t) = A
A = A * sin(w*t)
1 = sin(w*t)
w*t = asin(1) = pi/2
Plug in numbers
36.0 rad/s * t = pi/2
t = pi / (2*36.0 ) = 0.0436 s
minimum length y(t) = -A
-A = A*sin(w*t)
-1 = sin(w*t)
w*t = asin(-1) = 3/2*pi
36.0*t = 3/2*pi
t = 0.131 s