A spring force constant 210 N/m vibrates with an amplitude of 28.0 cm when .250 kg hangs from it.

What is the equation describing this motion as a function of time? Assume that the mass passes through the equilibrium point with positive velocity at t=0.

At what times will the spring have its maximum and minimum extensions?

To find the equation describing the motion of the spring as a function of time, we need to use the equation for simple harmonic motion:

x(t) = A * cos(ωt + φ)

where:
x(t) is the displacement of the mass from its equilibrium position at time t,
A is the amplitude of the motion,
ω is the angular frequency of the motion,
t is the time,
and φ is the phase constant.

First, let's find the angular frequency (ω):
ω = √(k / m)

where:
k is the spring constant (210 N/m),
m is the mass hanging from the spring (0.250 kg).

ω = √(210 N/m / 0.250 kg) = √(840 N/kg) = 28.99 rad/s (approximately)

Now, let's find the phase constant (φ):
Using the given information that the mass passes through the equilibrium point with positive velocity at t=0, we can use the formula:

φ = -π/2

Now we can plug in the values to get the equation describing the motion of the spring as a function of time:
x(t) = 0.28 * cos(28.99t - π/2)

To find the times when the spring has its maximum and minimum extensions, we need to find when the cosine function reaches its maximum and minimum values.

The maximum extension occurs when cos(28.99t - π/2) = 1.
Solving for t, we get:
28.99t - π/2 = 0
28.99t = π/2
t = π/2 / 28.99 ≈ 0.0545 seconds (approximately)

The minimum extension occurs when cos(28.99t - π/2) = -1.
Solving for t, we get:
28.99t - π/2 = π
28.99t = π + π/2
t = (π + π/2) / 28.99 ≈ 0.137 seconds (approximately)

Therefore, the spring has its maximum extension at around 0.0545 seconds and its minimum extension at around 0.137 seconds.

jkbb