physics
posted by sunny on .
A horizontal spring mass system is set up to keep time so that it goes from one side to the other in 6 seconds. The mass used in this system is 0.69 kg. The distance from one side to the other is 2.4 m.
What spring constant should be used if it is to oscillate correctly?
k = N/m
An equation of the form x(t)=Xmax cos(ωt) can be used to describe the motion of this springmass system. From the data given, determine the values of Xmax and ω.
Xmax = m
ω = rad/sec
An equation of the form v(t)= Vmax sin(ω2t) can be used to describe the velocity of this springmass system. From the data given, determine the values of Vmax and ω2.
Vmax = m/s
ω2 = rad/sec
An equation of the form a(t)=amax cos(ω3t) can be used to describe the acceleration of this springmass system. From the data given, determine the values of amax and ω3.
amax = m/s2
ω3 = rad/sec

period T = 12 seconds
amplitude = 1.2 meter
w = 2 pi f = 2 pi/T = sqrt (k/m)
2 pi/12 = sqrt (k/.69)
.5236 = sqrt(k/.69)
.2742 = k/.69
k = .189 Newtons/meter
w = .5236 radians/second we know already
Xmax = 1.2 meter, we know that too
so
x = 1.2 cos (.5236 t) meters
w = w2 = w3
the frequency of v and a is THE SAME as for the position!
then
v = 1.2*.5236 * sin(.5236 t)
so Vmax = .6283 m/s
a = 1.2 * .5236^2 cos (.5236 t)
=  w^2 x
amax = 1.2*.5236^2 =  .329 m/s^2