A vertical spring with spring stiffness constant 305 N / m oscillates with an am
ID: 2144085 • Letter: A
Question
A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of28.0 cm when 0.235 kg hangs from it. The mass passes through the equilibrium point (y=0) with positive velocity at t=0. Positive direction of y-axis is downward.
A) What equation describes this motion as a function of time? (Express your answer in terms of
t, where t is expressed in seconds.)
B) At what time will the spring stretch to its maximum length at first time? (Express your answer with the appropriate units.)
C) At what time will the spring shrink to its minimum length at first time? (Express your answer with the appropriate units.)
Explanation / Answer
A. y = Asin(omega*t) = 0.028*sin(34.25*t)
B. Solve the above equation for t when y = +0.028 :
For y = +0.028:
+1 = sin(36.02*t)
=> arc sin(+1) = 36.02*t
=> pi / 4 = 36.02*t
=> t = pi/(4*36.02)
=> t = 0.021 sec
C. Solve the equation for t when y = -0.028:
For y = -0.028:
-1 = sin(36.02*t)
=> arcsin(-1) = 36.02*t
=> 3pi/4 = 36.02 *t
=> t = 3pi/(4*36.02)
=> t = 0.065 sec