Patm = 1.013 x 105 Pa (H2O) = 1000 kg/m3 A spring expands a distance 0.25 m when
ID: 1967284 • Letter: P
Question
Patm = 1.013 x 105 Pa (H2O) = 1000 kg/m3A spring expands a distance 0.25 m when a force of 15.0 N is applied to it.
Suppose a block of mass 0.60 kg subsequently is attached to this spring and the
system of block and spring is set into simple harmonic motion horizontally with an
amplitude of 0.17 m. Ignoring friction, calculate
a) the angular frequency of the motion;
b) the period of the motion;
c) the maximum speed of the block;
d) the maximum acceleration of the block;
e) and find the length of a simple pendulum with the same period.
Explanation / Answer
F =kx k =F/x =15/0.25 =60 N/m A)k = force/distance = 15.0 /0.25 = 60 N/m Angular frequency w [omega] for a mass-spring oscillator is just given by w = sqrt(k/m) = sqrt(60/0.6) = sqrt(100) = 10 radians/s w =vk/m =v60/0.6 = 10 rad/s A =0.17 b) T = 2p/w T = 0.628 secs C) Vmax =Aw =0.17*10 = 1.7 m/s