Question

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?

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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