A vertical spring with spring stiffness constant 305 N/moscillates with an ampli

Question

A vertical spring with spring stiffness constant 305 N/moscillates 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.

Part A

What equation describes this motion as a function of time? Express your answer in terms of t, where t is expressed in seconds. '

Part B

At what time will the spring stretch to its maximum length at first time? Express your answer with the appropriate units.

Part 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) 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

B) 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

C) 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

Just a note: The way I vision this problem is the mass starts moving down at first so you will hit the longest section first and then the smaller section 2nd. If the mass is moving up first, then down, the answers would be flipped for b and c.

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