A 4.40kg block hangs from a spring with spring constant 2160N/m . The block is pulled down 4.60cm from the equilibrium position and given an initial velocity of 1.40m/s back toward equilibrium.
A) What is the frequency of the motion?
B) What is the amplitude?
C) What is the total mechanical energy of the motion?
let x = a sin (w t + p)
then
dx/dt = v = a w cos (w t + p)
d^2x/dt^2 = A = -a w^2 sin (wt+p)=-w^2 x
then
F = mA
-kx = m (- w^2 x)
w^2 = k/m
w = sqrt (k/m)
when w t = 2 pi, we have period T
so
w/(2 pi) = 1/T = f the frequency
so
f = (1/2pi) sqrt(k/m)
so
A. f = (1/2pi)sqrt (2160/4.4) = 3.53 Hz
then
at t = 0, x = -.046 and v = +1.4
w = 2 pi f = 22.2 radians/s
x = -.046 = a sin p
v = 1.4 = 22.2 a cos p
solve those for a and p
a = -.046/sin p
1.4 = 22.2 (-.046/sin p) cos p
so
tan p = -22.2(.046)/1.4
= -.729
p = -36 deg = -.63 radians
then a = -.046/sin p = .0783 meters
= 7.83 centimeters (that is answer B)
max velocity = w a when potential energy is 0
max energy = (1/2) w^2a^2