Consider a spring hung vertically from the ceiling. When a 2kg mass is attached
ID: 1511052 • Letter: C
Question
Consider a spring hung vertically from the ceiling. When a 2kg mass is attached to the spring, the spring is stretched 0.10m. What is the force constant of the spring? The 2 kg mass is removed and a different one attached to the spring. It then undergoes simple harmonic motion with a period of 2/5 seconds. What is the inertia of the new mass? If you pull the mass in part (b) so it undergoes simple harmonic motion with an amplitude of 10cm, what are its maximum acceleration and velocity? Taking upward to be positive, at what position in the motion do the maximum position, velocity and acceleration occur? If we call the time where it is first passing through the equilibrium position on the way up t = 0, write an equation of motion for the oscillation, x(t) =?, identifying the values of all constants that you use.Explanation / Answer
a)
weight is balanced by spring force
mg = kx
20 = k * 0.1
k = 200 N/m
b)
T = 2*pi*sqrt(m/k) = 2/5
m = 0.81 kg
c)
max acceleration when it experiance max spiring force so that will occure at extreams
kA = m*a_max
200 * 0.1 = 0.81 * a_max
a_max = 24.7 m/s^2
max velocity when there is zero forse acting on it
so at the equivlibrium position
so from energy balance at extream and equivlibrium position
kA^2/2 + kx^2/2 = kx^2/2 + mv^2/2
where x = streached position at equivlibrium position
kA^2 = m*(v_max)^2
v_max = sqrt(200 * 0.1^2 / 0.81) = 1.57 m/s
d)
assuming equivlibrium position to be zero reference
max position at 10cm up from zero
max vvelocity will at zer0 or at equivlibrium position directed up
max acceleration at the lowest extream = (10cm down to zero) directed up
e) by assumption in privicious part
x(t) = A*cos(wt + phi)
where A= 0.1 m
and at t=0 x=0
so
0 = 0.1 * cos(phi)
phi = - 90 degree
x(t) = 0.1 * sin(wt)
where w = 2 * pi * f = 2*pi /T
T = 2*pi*sqrt(m/k) = 2*pi*sqrt(0.81 / 200) =0.4 s
w = 15.7 rad/s
x(t) = 0.1 * sin(15.7 * t)