A mass of 1 kg is hung vertically from a spring stretching t

A mass of 1 kg is hung vertically from a spring stretching the spring 20 cm. It is then pulled 10 cm below its equilibrium position and released from rest. Find the angular frequency of oscillation the spring constant of the spring the amplitude of oscillation, the kinetic energy of mass when passing through equilibrium, the potential energy of the spring when the mass passes through equilibrium the time to first reach equilibrium the time to make one complete cycle.

Solution

let spring constant be k.

mg=kx

1 *9.8 =k*20*10^-2 so K =49 N/m

a) angular frequency = (k/m)^1/2 = (49/1)^1/2 = 7 rad per sec

b) we have already calculated k = 49 n/m

c)amplitude would be .1

d) at x=0 kinetic energy wouldbe max = 1/2 Kx^2 = 1/2 * 49 *(20*10^-2)^2 = .28 J

e) it would be 0

g) Time period = 2*pi (m/k)1/2 = 2*3.14 *(1/49)^1/2 = .897 second

f) time to first reach equilbrium would be T/4 = .897/4 =.224