An Ideal Spring Physics Problem?

Question

An ideal spring with a spring constant of 18 N/m is suspended vertically. A body of mass 0.60 kg is attached to the unstretched spring and released.

What is the extension of the spring when the speed is a maximum?

What is the maximum speed?

Answer 1

from energy
mgx=1/2 kx^2 so maximum extension x=2.m.g/k

x=0,653 m then amplitude r=x/2 ==> r=0,327 m


Vmax=w.r ==> w=2.pi/T T=2.pi.sqrt(m/k) T=1,147 s

Vmax=w.r=1,791 m/s extension is x= 0,327 m

Answer 2

At maximum speed the spring is at its static deflection extension, g*m/k. The maximum speed can be found by multiplying the amplitude of the displacement from static deflection (also = g*m/k) by the natural frequency in rad/s of the spring-mass system, sqrt(k/m).

Answer 3

The tension appearing on the spring, F = mg = 0.60 x 9.8 N we've, F = kx as a result, extension,x = F/ok = 0.60 x 9.8/15 = 0.39 m means power, U = (kx^2)/2 = 15 x (0.39^2)/2 = 1521 J the cost is optimum on the unstretched place of the spring. this is while the extension is 0. the means power on the extension 0.39 m is switched over into the kinetic power on the equilibrium place. i.e (mv^2)/2 = 1521 J v = sq. root of (2 x 1521/m) = (2 x 1521/0.60) = seventy one ms^-one million