WHEN A MASS IS ATTACHED TO A VERTICAL SPRING,THE SPRING IS STRETCHED A DISTANDE d.THE MASS IS THEN PULLED DOWN FROM THIS POSITION ANDS RELEASED.IT UNDERGOES 50 OSCILLATIONS IN30SECONDS.WHAT WAS THE DISTANCE d?
Spring constant = k = M*g/d
Oscillation frequency = 50/30 = 1.667 Hz
= [1/(2*pi)]*sqrt(k/M)
= [1/(2*pi)]*sqrt(g/d)
= 0.1591*sqrt(9.8/d) = 1.667 Hz
sqrt(9.8/d) = 10.48
9.8/d = 109.8
d = 8.9*10^-2 m = 8.9 cm