Question

A steel sphere of mass 0.02 kg attains a terminal speed vi = 0.5 m/s when dropped into a tall cylinder of oil. The same sphere is then attached to the free end of an ideal vertical spring of spring constant 8 N/m. The sphere is immersed in the same oil and set into vertical oscillation. Find 

(i) the damping constant 

(ii) the angular frequency of the damped SHM. (iii) Hence, write the equation for displacement of the damped SHM as a function of time, assuming that the initial amplitude is 10 cm. [g = 10 m/s²]

Answer 1

Data : m = 0.02 kg, v = 0.5 m/s, k = 8 N/m, 

A = 10 cm = 0.1 m, g = 10 m/s² 

When the sphere falls with terminal velocity in oil, the resultant force on it is zero. Therefore, the 

The equation of motion of the damped oscillation is resistive force and its weight are equal in magnitude and opposite in direction.

∴ |F_r| = βv_t = mg

where β is the damping constant.

∴ β = \(\frac{mg}{v_1}\) = \(\frac{0.02\times10}{0.5}\) = 0.4 kg/s

The angular frequency of the damped oscillation in oil,

The equation of motion of the damped oscillation is

x = Ae^(β/2m)t cos(w’t + φ) 

∴ x = (0.1 m)e^-(0.4/004)t cos (17.32t + φ) 

x = (0.1 m) e^-10t cos(17.32t + φ)