Question

At an outdoor market, a bunch of bananas attached to the bottom of a vertical spring of force constant 16.0 n/m is set into oscillatory motion with an amplitude of 20.0 cm. it is observed that the maximum speed of the bunch of bananas is 43.4 cm/s. what is the weight of the bananas in newtons?

Answer 1

Force Constant K = 16.0 N/m 
Amplitude of the motion = 20 cm = 0.2 m 
Speed of the Bananas = 43.4 cm/s = 0.434 m/s 
Gravitational Acceleration g = 9.8 m/s^2 
Calculating Angular velocity w = V / A = 0.434 / 0.2 = 2.17 rad/s 
We also have K = m x w^2 => m = K / w^2 = 16 / 2.17^2 = 3.39 kg 
Weight of the bananas in Newton = m x g = 3.39 x 9.81 = 33.33 N