Anybody know Oscillations for 10 pts a thorough answer.?

Question

A 2.00 kg block hangs from a spring. A 300 gram object hung below the block stretches the spring an additional 2.00 cm. The 300 gm object is removed and the 2.00 kg block begins oscillating. What is the period of oscillation?

Answer 1

First, the spring constant (I suppose g=10m/sec^2):
f=k.d
0.3 x 10=k x 0.02 so k = 150 N/m
Now, the problem is like that we have a 2 Kg block hung from the spring and pull it down 2 cm then release it. The total force applied to block would be:
f=Mg-kx=20-150x (x = distance of block from its final point, 2 cm above its first place) so the acceleration 'a' is:
a=10-75x so applying the second law:
x''=10-75x so the movement equation is x''+75x-10=0
Solving this differential equation gives sin function for x as the general form x=a.Cos(w.t+c). By substitution we get:
w^2=75 so w=5.sqrt(3) rad/sec
(divide it by 2*pi if you like to have the frequency in Hz)
T=2*pi/w ~ 0.73 sec

Answer 2

The perioid of oscillation will begin at the timeline,or point on your time line, when the 300gm object is being removed.. The period ends when the 2.kg block returns fully to the unextended position from which it started.

Answer 3

The period of free vibrations about equilibrium position is given by the formula

T = 2pi x square root of m/ k; where m is the mass and k is the restoring force constant.

0.3 x 9.8 newton force pulls it through 0.02 m

Therefore the force constant is 147 N/m.

T = 2pi x square root of 2/ 147 = 0.733 second.