Physics spring problem...extension and maximum speed.?

Question

An ideal spring with a spring constant of 15 N/m is suspended vertically. A body of mass 0.60 kg is attached to the unstretched spring and released. a.) what is the extension of the spring when the speed is a maximum? b.) What is the maximum speed?

Answer 1

a) This occurs when the force is equal to mg, so the extension is y = F/k = mg/k = .6*9.8/15 = .392 m

b) Vmax = A*ω where A = .392 m and ω = √(k/m) = 5 rad/sec.
Ergo, Vmax = .392*5 = 1.96 m/s

Answer 2

Maxspeed Springs

Answer 3

The force acting on the spring, F = mg = 0.60 x 9.8 N
We have, F = kx
therefore, extension,x = F/k = 0.60 x 9.8/15 = 0.39 m
Potential energy, U = (kx^2)/2 = 15 x (0.39^2)/2 = 1521 J
The speed is maximum at the unstretched position of the spring. That is when the extension is zero.
The potential energy at the extension 0.39 m is converted into the kinetic energy at the equilibrium position.
i.e (mv^2)/2 = 1521 J
v = square root of (2 x 1521/m) = (2 x 1521/0.60) = 71 ms^-1

Answer 4

in general
y(t)= A sin(wt + phase angle)
w=sqrt(k/m)
k - spring constant
m - mass

v(t)=dy/dt= Aw cos(wt + phase angle)
Vmax= wt + phase angle=0,pi,2pi...npi
t=phase angle/w