Question

I would like a step by step solution and clear to read please thank you.

2. A sphere of wood with a density of 720 kg/m', and radius- 1.2 cm is at the end of a 8 cm long cylinder of the same wood with a radius of 4 mm. It is placed in a beaker of water and, once it has reached equilibrium, it is lifted 2 cm and released, after which it bobs up and down with a decreasing amplitude. Set up theexpession for the bobber and determine the frequency of the oscillator and the expression for the height of the top of the bobber as a function of time. If the amplitude of motion decreases by 5% each cycle, determine the damping coefficient in e

Answer 1

density of wood = rho_wood = 720 kg m^-3 density of water = sigma_water = 1000 kg m^-3 length of apparatus = (L) (L) = 2r + 8 = 10.4 cm L = 10.4 times 10^-2 m T = 2 pi squareroot rho_wood L/sigma_water g = 2 times 3.14 squareroot 720 times 10.4 times 10^-2/1000 times 9.8 = 2 times 3.14 times 0.0814 T = 0.5488 second T = 0.55 second f = 1/T = 1.82 Hz Amplitude by 5% by dumping forces A = A_0 e^-t/T A = 95% of A_0 therefore 0.95 A_0 = A_0 e^-T/T 0.95 = e^-0.55/T 100/95 = e^0.55/T log (100/95) = 0.55/T 0.0513 = 0.55/T therefore T = 0.55/0.0513 = 10.72