F = kx
10 newtons = k * 0.25 meters
so
k = 40 Newtons/meter
Now do the problem
x = -0.15 at t = 0 and v = 0
so assume x = -0.15 cos (2 pi t/T)
then v = 0.15(2 pi/T) sin (2 pi t/T)
and
a = 0.15 (2pi/T)^2 cos (2pi t/T) = -(2 pi/T)^2 x
max F = m * max a
k (0.15) = m (2 pi/T)^2 (0.15
(2 pi/T)^2 = k/m = (40 Newtons/meter )/0.150 kg
f = 1/T
so
(2 pi f )^2 = 40/0.150
Now put one minute (t = 60 seconds) into the x equation
x = -0.15 cos (2 pi t/T)
A 150 gram mass is attached to a horizontally aligned spring on a frictionless surface. A force of 10 Newtons will stretch the spring 25 centimeters. If the spring is compressed to 15 centimeters and then released, calculate: the spring constant; the frequency and period of the system; and the position and speed of the mass one minute after it is released.
1 answer
1 answer