Question

Can someone please help with 1) part A of top and 2) part B of bottom? Can't seem to get them!

Part a A 0.561-kg ball is hung vertically from a spring. The spring stretches by 5.38 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will t take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position? Hints: This question contains more information than necessary. Also, beware of unit conversions! Enter answer here seconds

Answer 1

Mass = m = 0.561 kg stretched length at equilibrium = Delta x = 5.38 times 10^-2 m We know that at equilibrium mg = K Delta x so, k = spring constant = mg/4x = 0.561 times 9.81/5.38 times 10^-2 = 102.3N/m so, T = Time period = 2pi squareroot m/k = 2pi squareroot 0.561/102.3 = 0.465 seconds so, Total time = 4 times T = 1.86 seconds Ans Part B Delta x = stretched length at equilibrium = mg/k = 14.1 times 9.81/660 = 0.2096 m so, compressed Amplitude = A = 2 delta x = 2 times 0.2096 = 0.4192 m Ans