A 1.00-kg glider attached to a spring with a force constant 25.0 N/m oscillates

Question

A 1.00-kg glider attached to a spring with a force constant 25.0 N/m oscillates on a frictionless, horizontal air track. At t = 0, the glider is released from rest at x = -2.70 cm (that is, the spring is compressed by 2.70 cm). (a) Find the period of the glider's motion. (b) Find the maximum values of its speed and acceleration. speed m/s acceleration m/s^2 (c) Find the position, velocity, and acceleration as functions of time. (Where position is in m, velocity is in m/s, acceleration is in m/s^2, and t is in s. Use the following as necessary: t.) x(t) = v(t) = a(t) =

Explanation / Answer

(a) Find the period of its motion.

period T = 2(m/k) = 2(1.00kg / 25.0kg/s²) = 1.256 s

(b) Find the maximum values of its speed and acceleration

= 2 / T = 2 / 1.25s = 5 rad/s

max V = A = 2.70cm * 5rad/s = 13.5 cm/s
max a = A² = 2.70cm * (5rad/s)² = 67.5 cm/s²

(c) Find the position, velocity, and acceleration as functions of time (t).

Since the glider is at maximum amplitude at t = 0, model with the cos function:
x(t) = -Acos(t) = -2.70cm * cos(5t) for t in seconds
v(t) = dx/dt = Asin(t) = 13.5cm/s * sin(5t)
a(t) = dv/dt = A²cos(t) = 67.5cm/s² * cos(5t)

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