A vertically hanging Spring loaded with a Mass The upper end

A vertically hanging Spring loaded with a Mass: The upper end of a masless spring of constant k is attached to a fixed point. A box of mass m is hung at the lower end of the spring. Initially, you hold this box at rest with your hand at a vertical position so that the spring is neither stretched nor compressed. For all the processes described below, the spring remains completely elastic, Please set up a vertical y-axis pointing up with this resting position as the origin.

Solution

For Equilibrium position,

Fnet = 0

Fnet= ky - mg = 0

y = mg / k

so y position will be - mg / k .

b) As block will go down, gravitational PE will convert into KE and spring PE until

KE will becomes zero again.

Using energy conservation,

initial = final total energy

m g y + 0 + 0   = 0 + k y^2 / 2 + 0

y = 2 mg / k

hence y position = - 2 mg / k

c) now using energy conservation for inital and equilibrium position,

m g ( m g / k ) + 0 + 0 = 0 + k ( mg/k)^2 /2 + m v^2 /2

2 m g^2 / k = m g^2 / k + v^2

v^2 = m g^2 / k

v = g sqrt(m / k)

d) y(t) = A cos(wt)

m w^2 = k

w = sqrt (k /m )
y(t) = (mg / k ) coswt

where w = sqrt(k/m)

e) w = 2pi / T   = sqrt (k / m )

T = 2 pi sqrt(m/k)

time take to reach eq. to lowest position = T/4 = pi sqrt(m/k) / 2