Question

A 2.3 kg object oscillates at the end of a vertically hanging light spring once every...

A 2.3 kg object oscillates at the end of a vertically hanging light spring once every 0.40 s .

Write down the equation giving its position y (+ upward) as a function of time t. Assume the object started by being compressed 19 cm from the equilibrium position (where y = 0), and released. Note: the equilibrium position is defined here as that location of the mass at rest when it is freely hung from the spring, not the unstretched position of the spring with no mass.

How long will it take to get to the equilibrium position for the first time?

What will be its maximum speed?

What will be the object's maximum acceleration?

Homework Answers

Answer #1

a). Y(t) = Yo*cos(ωt) , ω = 2π/T = 2π/0.40s = 15.71 rad/s

Y(t) = (0.19 m) cos(15.71 t)

b). Y(t) = 0m ==> 0.19cos(15.71t) = 0

cos(15.71t) = 0 ==>15.71t = π/2 ==> t = 0.10 s

c). V(t) = Y'(t) = - ωYosin(ωt)

V(t) = - 15.71*0.19sin(15.71t)

V(t) = - 2.98 sin(15.71 t) , this expression gets it's maximum value when sin(15.71t) = ±1

Vmax = ± 2.98m/s

d). a(t) = V'(t) = - ω²Yocos(ωt)

a(t) = - (15.71)²*0.19cos(15.71t)

a(t) = - 46.9 cos(15.71t)

the acceleration is maximum when cos(15.71t) = 1 [the maximum value for the cosine function is 1] . a(max) = - 46.9m/s² and that firstly attained when t=0s and Y = Yo = 0.19m .

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